Comprehensive Guide to Periodic Motion and Spring-Mass Oscillations

So what exactly is periodic motion? Periodic motion is motion that repeats itself or that oscillates back and

forth. Some good examples of periodic motion and simple harmonic motion is the mass spring system and also the simple

pendulum. But let's talk about the mass spring system. So let's say this is the wall and we have a spring connected to a

mass horizontally and let's say that this is the equilibrium position of the spring.

So the spring has been stretched towards the right. So that means we're applying a force

to pull it towards the right to stretch it from its equilibrium position. As we pull it towards the right,

there is another force that wants to push it back towards its equilibrium position. This is known as the restoring

force. Now once you let go, the spring is going to move back towards

the left. Now it's not going to stop at the equilibrium position. It's going to move past it.

And once that happens, it's going to bounce back the other direction.

So it's going to move back and forth towards the left and towards the right and as it oscillates back and forth

you have simple harmonic motion or periodic motion. Now according to Hook's law the

restoring force is equal to * K * X.

X is the distance measured from the equilibrium position. So let's say this

is the wall and we have a spring and here's the block of mass m. The distance between where the spring is

located and the equilibrium position is x. At the middle at the equilibrium position x is zero. So the restoring

force will always be zero at the equilibrium position. K is known as the spring constant and K

has the units newtons per meter. So let's say if you have a spring constant of 100 newtons per meter, what

that means is that it requires 100 newtons of force to either stretch or compress the spring by 1 meter from its

equilibrium position. So if you wish to stretch it by 2 m, it requires 200 newtons of force.

So if you have a very stiff spring that's hard to compress, the K constant is very high. Let's say if K is a

th00and. So for this spring, it requires 1,000 newtons of force just to stretch it by

one meter. So this spring is going to be very stiff. It's hard to stretch or compress.

This particular spring is going to be easy to stretch or compress compared to the other one. So, this spring is

relatively loose. By the way, before we work on some practice problems, I do want to mention

one thing. Any equation in Hook's law, why is there a negative symbol for the restoring force?

Why do we have this negative sign? What's the reason for that? So notice that if you apply a force

towards the right to stretch the spring, the displacement vector will be towards the right and the restoring force is

towards the left. Because the restoring force is opposite to the displacement vector,

F_sub_R is negative because it's always opposite to the direction of motion. The restoring force will always act in

such a way to bring the object back to its equilibrium position. So if the equil position is here, the restoring

force wants to bring it to the left. Likewise, Let's say if the spring is compressed,

the restoring force wants to uh move it towards the right back to its equilibrium position. And

remember at the equilibrium position, the restoring force is zero since x is equal to zero and f_sub_r is kx. So k *

0 is zero. Let's work on some practice problems. How much force is required to stretch a

300 Newton per meter spring by 25 cm? So let's use the equation F is equal to kx.

And let's not worry about the negative sign. If you see the unit newtons per meter,

that tells you that that value represents the spring constant, which is 300 newtons per meter. And since it has

the unit meters in it, that tells us that we need to convert centimeters into meters.

It turns out that 100 cm is equal to a single meter. So to convert it we need to divide by

100 or move the decimal two units to the left. So this is equal to 0.25 m which

is the value of x. So what is 300 * 0.25 or what is 25% of 300?

This is going to be about 75. So that is the force required to stretch the spring by 25 cm.

How far can you compress the spring with a force of 150 newtons? Now let's think about this. It required

75 newtons to stretch the spring by 25 cm. It also requires 75 newtons to compress

the spring by 25 cm. So if we double the force to 150, the amount that we can stress, I mean

stretch or compress the spring is going to be twice the value. It's going to be 50 cm.

F= kx. If you double the distance, if you stretch it by two times the amount, the

force will increase by a factor of two. Likewise, if you double the force, you can double the distance that you stretch

or compress the spring bar. Now, for those of you who prefer to use an equation,

you can use this. Let's divide f_sub_2 by f_sub_1. f_sub_1 is equal to k * x1 and f_sub_2

is k * x2. Since we're dealing with the same spring, K is the same. So, as you can see, the force is

proportional to x. Let's say f_sub_1

is 75 newtons and we're looking for f_sub_2. X1. The distance that corresponds to 75

is 25 cmters. Actually, we're not looking for F2. We have F2. F2 is 150.

We're looking for X2. 150 / 75 is 2. And at this point, we can cross

multiply. So x2 is equal to 2 * 25 which is 50. Here's another problem. It takes 200

newtons to stretch a spring by 40 cm. How much force is required to stretch a spring by 120 cm?

So 40 cm corresponds to a force of 200 newtons. What is the force that's required to

triple the distance by a factor of three? If you triple the distance by a factor

of three, the force is going to increase by a factor of 3. So the force that's required is 600 newtons. Again, you can

use this equation. F_sub_2 over F_sub_1 is equal to X2 / X1. So we're looking for F_sub_2 this time. f_sub_1 is 200

newtons. X1 is 40 cm and X2 is 120 cmters. You don't need to convert the units to

meters because the unit centimeters will cancel. So 120 / 40 is 3.

So therefore f_sub_2 / 200 is equal to 3. And if we multiply both sides by 200, we can see that F_sub_2 is going to be 3

* 200, which is 600 newtons. So that's how much force is required to stretch a spring by 120

cm. Now what about the second part to the question? What is the value of the

spring constant? So using the formula F is equal to Kx. Solving for K, K is equal to the force /

the spring constant. So we can use the force of 200 newtons. And instead of using 40 cm, we need to convert it to

meters. So that's about 4 m. 200 /4

is equal to 500. So the spring constant is 500 newtons per meter. So to stretch the spring by a distance of 1 meter

requires a force of 500 newtons. Now what about the last part of the problem? How much work is required to

stretch a spring by 120 cm. Work is equal to force times displacement

which displacement could be represented as X. Now this is if you have a constant

force. Now what about if the force is a variable force? Example,

in this case, work is equal to the integral of f * dx, where f is a function of x.

And we know f is equal to k * x, ignoring the negative sign. So work is equal to the anti-derivative

of k * x dx. To find the anti-derivative of kx, add one to the exponent and divide by that

number. The anti-derivative of x^ the 1st power is x^2 / 2.

So the work required to stretch a spring is going to be basically 12 kx^ 2. This is also the potential energy that is

stored in a spring. In some textbooks, you'll see this U symbol. This is the elastic potential

energy of a spring. It's equal to 12 kx^2, which is the work required to stretch it

or compress it. So, k we know it's um 500 newtons per meter.

I'm going to write it like this. And x is going to be 120 cm, but we'll need to

convert it to meters. So, let's divide it by 100, which is 1.2 m squared. And the unit meters will cancel,

which leaves behind newtons time meters. There's 2 meters here. Since it's new time me is equal to the unit jewels.

Half of 500 is 250 * 1.2^2 is equal to 360. So it takes 360 jewels of work to

stretch this particular spring by 120 cm. Now let's go back to our mass spring

system slash oscillator. So let's say if we have a mass of m attached to a spring and let's say

this is the equilibrium position. Now, as we mentioned before, the force

required to stretch the spring is the apply force. And whenever you stretch it, there's

going to be a restoring force that pulls it back towards the left. Now, you also need to know that at this

point, the net force is zero. Now, this is only true while you're holding it. As soon as you release the

mass from rest, the applied force will disappear. At that point, the net force is equal to

the restoring force. And at this instant, as soon as you release it, this object has maximum

acceleration. And initially, the velocity is zero. The velocity reaches its maximum value

once it reaches equilibrium. At equilibrium, the restoring force is zero. And because the restoring force is

zero, the acceleration is equal to zero at the equilibrium position. However, because the velocity is at a maximum,

it's moving towards the left. So, because it has momentum, because it's moving due to inertia, it's going

to continue to move to the left, compressing the spring. As it begins to compress the spring,

a restoring force is going to build up, decelerating the object. Once the object reaches a velocity of

zero, the restor is at its maximum value, which means acceleration it's at its maximum value at this point

as well. So make sure you understand this that at the equilibrium position, the velocity is at its maximum and the

acceleration is zero. When it's fully stretched or compressed, the acceleration

is at its maximum value and the velocity is zero. Now, if no friction is present in a

system, this oscillator will continue to oscillate indefinitely. If there's friction, friction is going

to oppose motion. So, in the first picture, the object is moving towards the left. So the force of friction will

be directed towards the right. Now in the middle it's still moving towards

the left. So friction is still directed towards the right. And on the last picture because the velocity

is zero, friction will be zero since it's not moving instantaneously. But as it begins to move towards the right,

friction will oppose it. So friction always opposes motion. By the way, once you release it, there's

no friction until it begins moving towards the left. So, as it accelerates, friction is going to oppose it.

Whenever there's friction present, the oscillator will be dampened by it. The amplitude will gradually decrease.

Now, let's talk about X and A. at the equilibrium position. Let me draw a new picture.

So let's say if we have an oscillator and let's say this is the equilibrium position

x is equal to zero at the equilibrium position and let's say at this point x is equal

to one. So this is the maximum displacement. The maximum displacement is represented

by the symbol capital A. So in this case for this problem a is one. At this position X is.5

since it's to the right of the equilibrium position. And at this position X is.5.

So let's say K is 100 newtons per meter. What is the force the restoring force at let's say a position of.5

include the appropriate sign as well. So F is equal to kx it's going to be * 100 *.5

which is50. So at this point we have a force of -50. The negative

sign means that it's towards the left. Now what about at this position? If we plug in.5.

Okay, that does not look like a five. It's going to be *.5

which is pos 50. Since the object is moving at this location towards the left, the restoring

force is towards the right. And notice that the force is positive 50 because it's directed towards the right.

When the restoring force is directed towards the left, notice that it's negative.

Now, let's talk about the energy of this particular oscillator. When it's fully stretched and once you

release it, the acceleration is at its maximum and the velocity is zero. The kinetic energy of any moving object is

12 mv^ 2. So the kinetic energy is dependent on the speed. So when it's fully stretched or fully compressed,

let's say it's fully compressed at this position here, the kinetic energy is at a minimum because the velocity is

zero. So K is going to equal zero when it's fully stretched or compressed. Now at

the equilibrium position, the velocity is at a max. So therefore K is at its maximum value at the

equilibrium position. As we mentioned before the acceleration is zero at the equilibrium position and

the restoring force is equal to zero since X is zero and F= KX. So K * Z is zero.

Now what about the potential energy? The potential energy is at its maximum

when it's fully stretched or compressed. But at equilibrium, the potential energy is zero. The potential energy is equal

to 12 kx^ 2. Since x is zero at the equilibrium position, the potential energy is zero because 12 k * 0^ 2 is 0.

Now there's another thing that you need to be familiar with and that is mechanical energy.

The mechanical energy is the sum of the kinetic energy and the potential energy. Keep in mind

when the amplitude is one x can vary anywhere between -1 to 1. The mechanical energy depends on the

amplitude that is the maximum value of x or the maximum displacement. So the mechanical energy is 12 k a^ 2.

The kinetic energy is 12 mv ^2 and the potential energy is 12 kx^2. So whenever x is equal to a the

potential energy is the same as the mechanical energy at this point the kinetic energy is zero.

And whenever x is equal to zero that is at the equilibrium position potential energy is equal to zero. So

let me organize this information. So let's say this is the equilibrium position where x is zero.

At that point the potential energy will be equal to zero. So k is at its maximum. K equals the

mechanical energy. The mechanical energy of the system is constant if there's no friction.

The mechanical energy is the total energy of the system. Now, let's say if it's at the right, if

it's fully stretched at this point, the maximum displacement is equal to a.

So, when x is equal to a, the potential energy is at a maximum. it's equal to the mechanical energy which means that K

is equal to zero and when it's fully compressed P is still equal to the mechanical

energy and K is equal to zero. Now sometimes you may need to calculate the maximum acceleration and the maximum

velocity. What equations can we use to do that? So let's start with maximum velocity.

Now we said that it's going to be the velocity is going to be at its maximum when x is equal to

zero. So when x is equal to zero the potential energy is zero. So this disappears. So therefore 12 k a^2

is equal to 12 mv^2. We can multiply both sides by two to get rid of the fraction.

So k a^2 is equal to mv ^2. To isolate v ^2 we need to divide both sides by m.

So V is equal to the square root of K / M time the amplitude A.

So that's how you could find the maximum velocity if you ever need to. What about the acceleration? How can we

find a maximum acceleration? We know that f is equal to kx. And according to Newton's second law, F

is ma. So what we need to do is divide both sides by m. So k / m * x is equal to the

acceleration at any point. So that's the acceleration as a function of x.

Now the acceleration is at its maximum value when x is the maximum which is the amplitude.

So the maximum acceleration is k * a / m. Now what if we don't want to find the

maximum velocity? What if we need to find the velocity as a function of x? Let's say when x is 4.7. How can we do

that? Well, let's go back to our conservation of energy equation. Mechanical energy is equal to kinetic

plus potential energy. So, let's begin by multiplying everything by two.

2 * a half is 1. This will cancel all of the fractions. So we have k a^2 is equal to mv^2

+ kx^2. Now let's subtract kx^ 2 by both sides. k a^ 2us kx^2 is equal to m v^2. And now

let's factor out the gcf the greatest common factor which is k. So this is going to be a^2 - x^2.

Next, let's divide the expression by the mass. So, v ^ 2 is equal to k / m * a^ 2 -

x^2. Now, what's our next step? The next thing we need to do is take out

a I mean a^2 if we take out a^2 a^2 / a^2 is simply 1

and this / a^2 that's going to be x^2 / a^2. So now what we want to do is take the

square root of both sides. So the velocity as a function of x is going to be the square roo<unk> of k / m

* a* the square<unk> of 1 - x^ 2 / a^ 2. Now the velocity can be positive or

negative. If the velocity is moving towards the right it's positive. If it's moving towards the left it's negative.

Now if you recall the quantity<unk> km m * a is equal to the maximum velocity.

So therefore we can rewrite the equation like this. So the velocity as a function of x is equal to plus or minus the

maximum velocity times the square root of 1 - x^2 / a^ 2. So make sure you uh

write this equation down. It might be useful later. Now there's something else that we need

to talk about and that is the frequency and period of this spring mass oscillator.

So let's say this is the equilibrium position. Now as the spring

moves to the left and as it moves back to where it started from that is equal to one cycle.

The distance of one cycle is equal to basically four times the value of x.

So from here back to its equilibrium position, it travels a distance of x. And then when it's fully compressed, it

travels another distance of x. And then when it goes back to its equilibrium position, that's another x

value. And when it returns to where it started, that's X again. So the distance of one cycle is four times its maximum

displacement. So we should really say this is four * A since a represents the maximum displacement.

So this is a. Now what about the period and the frequency?

The period is the time it takes to complete one cycle. The period which is represented by

capital T is the number of it's the time divided by the number of cycles. So let's say if it takes 30 seconds

to complete 10 cycles, 30 / 10 is three. That means that it takes 3 seconds to complete a single

cycle. Therefore, the period is three. So that's how you can find the period. It's the total time divided by the total

number of cycles or the time it takes to complete one single cycle. The frequency is the number of cycles per second. So

if you take the total number of cycles and divide it by the total time, you can get the frequency.

So, for example, let's say if it takes, let me choose a nice number. Um, let's say if this system can cycle 200

times in 10 seconds. So, if it makes 200 cycles in 10 seconds,

that means that it's doing 20 cycles in 1 second. So, the frequency is 20 or 20 hertz. Hertz is the unit of

frequency. Now frequency and period are inversely related. The frequency is 1 / the period and the period is 1 / the

frequency. So let's work on some problems. What happens to the following if the maximum

displacement of the spring is doubled? So what happens to the total energy of the system? The total energy of the

system is the mechanical energy. Now the first thing we want to do is write an equation that relates the mechanical

energy with the maximum displacement and that's this equation. Me is equal to 12 k a^ 2 where a is the maximum

displacement. So if we double the value of a what happens to me for these types of

problems anything that's constant or remains the same replace it with a one. So 12 k we're just going to replace it

with one and plug in the value of a. If you double the value of a because the function is squared, the mechanical

energy will increase by a factor of four. If we were to triple the value of a, the mechanical energy will increase

by 3^ 2, which is 9. If we quadruple the value of a, the mechanical energy will increase by a factor of 16.

Now, what about part B? The maximum velocity. What equation can we write that relates maximum velocity to the

maximum displacement? Now we came up with the equation earlier. We said that the maximum velocity

is equal to the square root of k / m where k is the spring constant times the maximum

displacement. So notice that it's raised to the first power. So if we double the value of a

and this is going to stay the same. So we're going to replace it with one. The velocity will increase by a factor of

two. If you triple the displacement, the velocity will increase by a factor of three. If you cut the displacement in

half, the velocity will decrease by a factor of two or it's going to be 1/2 of its original value. Now what about part

C? The maximum acceleration. The equation that we wrote for that is this equation. The maximum acceleration is

equal to K / M times the maximum displacement. So if you double the value of the maximum displacement, the

acceleration will increase by a factor of two. If you triple it, the acceleration will triple. If you reduce

it by 1/2, the acceleration will reduce by a factor of two. Let's work on this problem. A massless

spring is pulled by 30 cm from its natural length and then released. The spring is attached to a4 kg block and

oscillates across a horizontal friction with surface. Let's begin by calculating the maximum

velocity. So if you want to you can draw a picture.

So here's the block and it rests across a horizontal surface and let's say this is the equilibrium position.

So right now it's pulled by 30 cm which is.3 m. So let's make a list of what we know. A

the maximum displacement is.3 and K the spring constant is 300 newtons per meter.

What equation do we need to calculate the maximum velocity? The maximum velocity is simply

the square root of K / M * A. The mass is4 kg. So we have everything we need to find the maximum velocity.

So this is going to be<unk> 300 /4 *.3 300 /4

is 750. The square root of 750 is 27.386 *.3

and you should get a maximum velocity of 8.216 216 m/s.

So now what about part B? What's the maximum acceleration? So let's use this equation. It's equal

to K * A / M. So it's going to be 300 *.3 / 04.3

/4 is basically 75 * 300 that's 225. So the maximum acceleration is 225 m/s squared.

Now what about part C? How can we find the velocity at a position of 20 cm from its natural

length? So, keep in mind the maximum velocity was about 8.2 something

8.216. We can use this equation. The velocity as a function of x is equal

to the maximum velocity times the square root 1 - x^ 2 / a^ 2.

That's the equation that we had earlier uh in this video. So we want to find the velocity at a position of 20 cm from its

natural length. Now, because we have a ratio between x and a, it doesn't really matter if you

use centimeters or meters. The units simply have to match because they're both going to cancel. Centimeters over

centimeters and meters over meters will cancel. So, we can plug in 20 cm if we want to.

So, the maximum velocity is 8.216. And then this is going to be 1 - 20^ 2 / 30^ 2

/ 30^ 2 is about 4 over 9 which is4 repeating and 1 minus that number is about.5 repeating and the square root of

that is 7454* 8.216. So, you should get a velocity of about

let's make some space. 6.124. So, your answer should be less than the

maximum velocity. If for some reason it's larger, you know it's not correct. Now, what about the next part? How can

we find the restoring force 20 cm from its natural length? Now let's say 20 cm is located at this

point. We know the restoring force is directed towards the left which means that it should be negative.

So the restoring force is equal to * kx. So this is going to be negative 300 newtons per meter

time x which is 2 m. Now in this case we need to convert centimeters to meters because we're using k and k is in

meters. 3 * 2 is 6. So 300 *2 must be 60. So the restoring force is 60 newtons.

It's negative because it's directed towards the left. Now, what about the acceleration?

Well, we know that the force is equal to mass time acceleration. So, the acceleration is the restoring force / m.

And the restor is -60 newton. The mass is4 kg. So -60 /4

is -150 m/s squared. Keep in mind the maximum acceleration was 225 and this is less than 225. So that works out.

Now you can also use this equation. Acceleration is equal to K / M * X.

It's similar to this equation. The maximum acceleration is K * A / M. But here it's K * X / M. So K is 300, X is20

and M is4. So if you use this equation, you should get the same answer

which is 150. Now we know it should be 150 because the resultant I mean the

restoring force is towards the left acceleration and the force that causes that

acceleration are usually in the same direction. So because the restoring force is negative the acceleration is

negative which really means that this equation should have a negative sign. So let's go ahead and add it.

So that is it for this problem. In this problem, we have a vertical spring.

So let's say this is the spring without a mass attached to it. Now once we add a mass to it, it's going to

stretch. Now it stretches by 4 m.

So this is the new equilibrium position of the mass. Now if we stretch it even further

by an additional 20 m from its new uh equilibrium position.

Once we release it, it's going to bounce up and down. So with this information

determine the spring constant, the amplitude and everything else. So let's focus on the spring constant.

We can use this picture to find the value of the spring constant. The applied force that stretches the

spring from its original position is basically the weight force of the object which is mg.

The spring wants to go back to its original length. So the restor

is equal to the weight force so that the net force is zero creating that new equilibrium position.

So f_sub_r is equal to mg and the restoring force is kx and g is 9.8. So you really don't need

to worry about the negative sign. They will cancel. Now let's solve for k. So k is going to be equal to mg / x. So the

mass is 2 kg. G is 9.8. 8 and it stretches by a distance of4 m. So 2 * 9.8 is 19.6 /4

is 49. So K is 49 newtons per meter.

Now what about part B? What is the amplitude in this problem? What do you think the amplitude is equal

to? Now remember this is the new equilibrium position and it's stretched 20 m from

that equilibrium position. So therefore the amplitude is 20. So once we release it from this position

it's going to bounce up and down 20 from this position. So it's going to go up 20 and down 20 m.

Now what about C? How can we calculate the maximum acceleration? What is the equation that we need? If

you recall the maximum acceleration is equal to the spring constant time the amplitude divided by the mass. So it's

49 *2 / the mass of 2. So the maximum acceleration is 4.9 m/s

squared. Now what about d the maximum velocity? The equation that we need for that

is the square root of k / m * a 49 / 2 is 24.5 and the square root of that

is about 4.95* *2 that's about.99

m/s. So now what about part E? What is the kinetic energy, potential energy and the

mechanical energy when it's 10 m from its equilibrium position? So let's find the

potential energy first. The potential energy is 12 kx^2 which is basically 12 * 49 times an x

value of10 49 *10^ 2 that's about49 * half that's 0.25.

So the potential energy is.245 Jew.

Now what about the kinetic energy? The kinetic energy is equal to 12 mv^2.

Now we can't use the maximum velocity. The velocity will be.99. It will be at its max when x is zero at the

equilibrium position. But we're not at the equilibrium position. X is 10. So we got to find the velocity first at 10.

The velocity as a function of X is the maximum velocity*<unk> of 1 - X^ 2 / A^ 2.

So the maximum velocity is about.99 and x is 0.1 a the amplitude is 2.1^ 2 /2^ 2 is 1/4 1 - 1/4 is 3/4s and the

square<unk> of 3/4s is about 866 times.99 you should get 8574

m/s. second. Now that we have the velocity, we can now use this equation to find the

kinetic energy. So let's plug it into that equation. So K is equal to 12 * the mass which is 2 kg

time the speed.8574 2. So half of 2 is 1 and 1 *85 74^ 2 is about 735 jew. So that's the

kinetic energy 735. Now let's calculate the mechanical

energy. We're going to do it two ways. So first let's use the equation

mechanical energy is equal to 12 K * the amplitude squared. So K is 49 and the amplitude

is 20 m or2 m2 is 1 over 25 or 04* 49 which is 1.96. Half of that is.98.

So the mechanical energy which represents the total energy of the system is.98 Jew. Another way in which

we could find it is by adding the potential energy and the kinetic energy. So if we add245 plus 735 it also gives

us the same answer of 98 uh jewels. So let's say if you didn't know the

equation to find the velocity at any point x, you can use this to get the mechanical energy. And if you subtract

the mechanical energy by the potential, that's one way you can get the kinetic energy.

So there's different ways of solving different things that you may need. So you're not faced with one option.

There's many options that you can use. Try this problem. A spring is compressed.35

m with a 0.25 kg block by an applied force of 500 newtons. What speed will the block have as soon

as it's released from the spring across a horizontal frictionless surface? So let's say this is the wall

and we have a compressed spring and the mass attached to it

and let's say this is the horizontal frictionless surface. Now it's compressed by an applied force

of 500 ntons and it's compressed by a distance let's say of.35

m and the mass of the block is 0.25. Right now the applied force is balanced

by a restoring force of 500 Ntons. So as soon as we release it the applied force will disappear and it's going to

accelerate towards the right due to this restoring force of 500. Now this restoring force is not

constant. As the spring expands, the restoring force will decrease eventually to zero.

At that point, the mass will be released from the spring and it's going to move to the right at constant speed. So, when

it's completely released from the spring, when it's no longer attached to the spring, what is the speed that the

block will have? How can we calculate that value? Take a minute and work on that example.

The first thing that we need to do is find the spring constant K. So F is equal to Kx. Therefore K is the

force / X. So we have an applied force of 500 newtons which compresses the spring by.35 m.

So if we divide these two, we'll see that the spring constant is about

14 29 rounded to the nearest whole number

newtons per meter. So now that we have the spring constant, we can use conservation of energy to calculate

how fast it's going to move. So the mechanical energy is equal to the kinetic energy

or you can say the potential energy stored in the spring is equal to the kinetic.

The mechanical energy is 12 k a^2. If you use the potential energy equation, it's 12 kx^2. But in this case, x is

equal to a because it's stretched. I mean, it's compressed all the way. X is basically at its maximum displacement

when it's.35. So we'll use the symbol a though. 12 k a 2 is equal to 12 mv^ 2.

So if we multiply both sides by two, we can get rid of the 12. So K is 1429,

A is.35, M is.25, and we need to solve for V.35^

2* 1429. That's about 175. And if we divide that

by 0.25, we can see that V ^2 is equal to about 700.21.

And now if we take the square root of both sides, the velocity is about 26.5

m/s. So that's how fast it's going to be moving.

By the way, let's say if we wanted to find the kinetic energy of the object

as soon as it's released using this equation, it's 12 times the mass, which is 0.25

times the speed, which is really 26.46 squared. The kinetic energy of the object is

about 87.53 Jew. Now another equation that you can use to

get that answer is the work equation. So keep in mind as soon as we release it, there's a restoring force that's going

to accelerate towards the right. Work is equal to force times distance. However, we don't have a constant force.

We have a variable force. Now, let's assume that this force decreases constantly. Well, it actually

does. So, we don't really have to assume. The force will decrease at a constant

rate from its maximum value of 500 to zero. as the block is displaced by.35 m.

So basically the energy transferred by this force should be equal to the kinetic energy

which is 87.5. So how can we find the area of that triangle?

Well area is basically 12 base time height or 12 * the height which is 500. That's

basically the restoring force times the base which is the displacement of.35. Half of 500 is 250 and 250 *.35

gives you 87.5 Jew. So that's another way in which you can calculate the kinetic energy gained by

the block is by using the work equation. But if you're going to use the restoring force, you need to add a 1/2 to it

because the restoring force is not constant. Now there are some other topics that you

need to know and that is the equations for the period and the frequency as it is related to the mass and the spring

constant of a spring. Now we need to compare circular motion with the simple harmonic motion of a

spring. So let's draw a circle. Now

let's say if an object on a circle moves from let's say point A to point B

as it moves this way. If you can view the circle horizontally from this perspective

rather than viewing it from uh top to bottom. If you can just have like a side view,

all you will see is that this particle is moving in the negative x direction. You're only going to see the

displacement along the x-axis. Now, as it moves from, let's say, position B to position C,

if you take a side view of it, you're going to see it's still moving towards the left.

Now, as it moves from C to D, it's going to appear as if it it's moving towards the right, which it

is. And from D to A, it's moving this way. So if you can view

it from the side, circular motion would seem similar to the oscillation of a spring as it moves back and forth.

So knowing that we can say that the speed of the object around a circle is

basically the distance that it travels divided by the time. And the distance around a circle is the circumference

which is 2 pi r. The time it takes to make one revolution or one cycle is the period which is

represented by capital t. Now notice that the radius of the circle is basically the amplitude.

So we can replace r with a. Now we no longer need this picture. So now at this point let's multiply both

sides by t. So vt is equal to 2 pi a. And now let's divide both sides by v. So the period is 2 pi * the amplitude /

the velocity. And if you recall the maximum velocity we said was equal to

the square roo<unk> of k / m * a. So to solve for a, we can multiply both sides by the square root of m / k. So

that m will cancel on the right and k will cancel on the right as well. So v * the<unk> of m / k is equal to a. So now

let's replace this a with this expression. So the period t is 2<unk>i / v times v

max time<unk> m / k. So we can cancel v. So therefore the period

of a spring is 2<unk>i * the square<unk> of m / k. So now let's talk about this equation.

If we increase the mass of the spring, what's going to happen to the period? Will the time it takes to complete one

cycle, will it increase or decrease? Because the mass is in the numerator of the fraction, the period will increase

as well. So let's say if you double the mass of the spring, what effect will it have on

the period? So for these questions, plug in the value that is being changed. Everything else, plug in the one. So if

we double the mass, the period will increase by the roo<unk> of two. Now what if we quadruple the

mass? The period will increase by the square roo<unk> of 4 which is two. So if you

quadruple the mass, the period will double. If you increase the mass by a factor of

9, the period will triple in value. What if you reduce the mass by 1/2?

the square root of 12 which is 1 over <unk>2 that's <unk>2 over2. So the period will increase well

actually it's going to decrease by <unk>2 over2 that's about 707. So it's going to be 70% of its value

or you could say it's reduced by a factor of <unk>2 because you're dividing it by <unk>2.

So if you decrease the mass to half of its value, the period is going to be reduced by a factor of <unk>2.

It's going to be 70% of its original value or 70.7. Now what about the spring constant?

Let's say if we increase the value of K, what happens to the period? The period will decrease since K is on

the bottom. So if you double the value of K, the period will decrease

by the square root of two. It's going to be 70% or 70.7% of its original value. If you quadruple the value of K,

the period is going to be 1/2 of its original value. Now what if you reduce the spring

constant by 1/4? What effect will that have on the period? 1 / 1/4 is 4. So if you decrease

K to 1/4 its value, the period will double. Now let's think about what this means.

So we could see from the equation why the period will increase if we increase the mass. But

let's understand it conceptually. As you increase the mass of the block, it's going to have more inertia. So it's

going to be harder to move. As a result, it's going to have less acceleration. F is equal to ma.

If the force is constant, whenever you increase the mass, the acceleration will decrease.

So if it has a lower acceleration, it's going to move more slowly. And therefore, an object that moves slowly

takes a longer time to get to its destination. Therefore, as you increase the mass, the period increases. Since

the object is heavier, it's going to take a longer time to to move back and forth. And so that's why the period

increases. Now what about the second situation where if we increase the value of K the

period decreases. A large K value means that the spring requires a greater force to compress or

stretch it. So the spring is stiff. Because the spring is stiff, it doesn't really move very far.

It doesn't really it's not easily stretched or compressed. So it can vibrate faster

as a result because it can vibrate faster its period will be reduced.

So keep in mind K is FX. So a large K constant means that a greater force is required to stretch it

with the same distance. And if the force is greater, the acceleration will be greater as well. And a higher

acceleration means a fast moving object. It can quickly gain speed. And if the object is moving faster, then it's going

to take a longer time, I mean a shorter time actually, to oscillate back and forth. And so that's why as you increase

the value of K, it oscillates faster and the period is reduced. Now going back to this equation,

there's one more equation that you need and that is frequency. Frequency is 1 / period. So the frequency is going to be

1 / 2<unk>i *<unk> k / m. So this time if you increase the K value, if you increase the spring constant,

a greater force is required to stretch or compress the spring, which means that there's going to be a greater

acceleration, which means that it's oscillating faster. And because it's oscillating faster, the frequency will

increase. Now if you increase the mass, the object requires

a greater acceleration to move it back and forth. But if the force is constant, as you increase the mass, the

acceleration will be reduced. And so it's going to be moving slower, so to speak. So therefore, the frequency will

be reduced. Frequency and speed, they're related. An object that can oscillate back and forth

at a faster rate will have a higher frequency. And an object that oscillates at a

slower rate will have a low frequency. So as you increase the value of K, the speed of the oscillations will increase

and so the frequency will increase. If you increase the mass, heavy objects tend to move slower and so the frequency

will decrease. A spring attached to a.25 kg block is stretched horizontally across a

frictionless floor. 25 m by an applied force of 200 newtons. Calculate the frequency and period of

the oscillations of this mass spring system. So let's begin by drawing a picture.

So here we have a spring attached to a mass and here is the floor.

So as we stretch it with an applied force once we release it the spring will

oscillate back and forth. So how can we find the frequency of the oscillations? Well we can use this

equation 1 / 2<unk>i * the square root of k / m. Now in this problem we don't have the

spring constant K. So we need to find it. But we do have the applied force and the distance that it's stretched by. So

K is equal to the force / X. So that's 200 newtons /.25 m. So let's go ahead and divide those

two numbers. So K is 800 newtons per meter. Now that we have the spring constant, we

can calculate the frequency. So it's 1 / 2 pi * the<unk> of k which is 800 / the mass which is 0.25 kg.

800 / 0.25 is 3200. And the square root of 3200 is about 56.57. And let's divide that by 2 pi.

and you should get a frequency of 9 hertz. Now that we have the frequency, we can

easily calculate the period. The period is simply 1 / f, which is basically 1 / 9.

1 / 9 is about 0.1 repeating. So the period is about1 seconds. So it takes1

seconds to make a full cycle. That is it takes 111 seconds for the block to travel towards the left and back to its

original position. And in 1 second the spring will oscillate nine times.

That's the frequency, the number of cycles that occurs in 1 second. So in 1 second it's going to go backwards and

forwards nine times in that single second. Here's another problem you can try. When

a 70 kg person gets on a 1200 kg car, the springs compress by 2 cm, what will be the frequency of vibration

when the car hits a bump? So, first let's find K. So, F is equal to Kx.

The force that compresses the spring by 2 centimeters is the weight of the person because once he gets on the car,

X changes by 2 cm. So we only need to use the weight of that person and not the weight of the car and a person to

find K. So K is equal to mg / X. So it's the mass of the person times 9.8 8 /

2 cm converted to meters. To convert centimeters to meters, divide by 100. So this is 02 m.

So it's 70 * 9.8 / that number. And the spring constant is 34,300 newtons per meter. Now that we have the

spring constant, we can calculate the frequency of vibration. Now when the car hits a bump,

you have to consider the mass of the person and the car because the spring is attached to both or it's affected by the

mass of the person and the car. It has to support the weight of both. So using the equation f is equal to 1 /

2 pi * the<unk> of k / m. This is 1 over 2 pi time<unk> 34,300

divided by the total mass. That's uh 70 + 1 1200 or 1270. 34,300

/ 1270 is about 27. And the square root of 27 is about 5.2.

And if you divide that by 2 pi, you should get a frequency of83 hertz.

So that's it for this particular problem. So here's another problem. Here we have

an insect caught in a spiderweb. We have the mass of the insect and the frequency and we need to find the spring constant

K. So let's start with the equation f is equal to 1 / 2<unk>i

*<unk> of k / m. Now let's solve for k.

So first let's multiply both sides by 2 pi. So 2 pi f is equal to the square root of

k / m. Now at this point we're going to square both sides so we can get rid of the

square root on the right. So 2 p^2 is equal to k / m. Now the last thing we need to do is multiply both sides by m.

So now we have the equation that we need. So the spring constant K is equal to the

mass * 2 pi F squar and let's assume that the mass of the spiderweb is negligible.

So we're only concerned with the mass of the insect. Now we need to plug in the mass in

kilograms. To convert grams to kilograms you can divide by a th00and or instead of writing 0.25g you can write 0.25. 25

* 10 - 3 kg. 1 g is about 0.1 kg or 1 * 10 - 3 kg. And the frequency is 20 hertz.

2 pi * 20 is about 40 pi or 125.66. If you square it, you should get 15791. And then if you multiply by 0.25 * 10us

3, this should give you a spring constant of 3.95

newtons per meter. Now that we have the spring constant K, how can we answer the second part of the

problem? So if the mass is now 10 g, what is the frequency? So

using the equation 1 / 2<unk>i *<unk> k m, it's going to be different.

So k is about 3 9478. 3.95 is the right answer. And the mass

is going to be.1 * 10us 3 kg. So 3.9478 /.1 * 10us 3 that's 39,478.

If you take the square root of that you should get 198.69 and then divide that by 2 pi

and this will be equal to 31.6 hertz. Now you can get the same answer even if

you don't have the value of k. And let me show you. So let's say if we had a ratio f_sub_2 /

f_sub_1. f_sub_2 is 1 / 2<unk>i square<unk> k / m2.

And we're assuming that k is the same since we're dealing with the same spider web. And this is going to be the same

thing but m1. So we can cancel the 2 pi and we can cancel k. So we get the equation f_sub_2 / f_sub_1

is equal to the<unk> of 1 / m2 / the<unk> of 1 / m1. So how can we simplify

the expression that we have on the right? What would you do to simplify it? Let's multiply the top and the bottom by

the square root of m1. If we do that, m1 will cancel on the bottom. And so we're going to have f_sub_2 /

f_sub_1. And if you multiply 1 over m2 * m1, this is just going to be the square root of

m1 over m2. So this is the equation that we can use to get the answer to the second part of

the problem if we don't have the spring constant k. So notice that a frequency of 20 hertz

corresponds to a mass of 25 g. Because we have a ratio of the masses, we don't need to convert it to kilogram.

So m1 is going to be 0.25 g. Now we're looking for the frequency f_sub_2 when the mass is10.25

/10 is 2.5. And the square root of 2.5 is about 1.581.

So f_sub_2 / 20 is equal to 1.581. To solve for f_sub_2, multiply both sides by 20. So 1.581 * 20

is 31.62 hertz, which is about the answer that we got a few minutes ago.

A block of mass attached to a spring with a constant of 200 newtons per meter vibrates at 15 hertz. What is the

frequency of vibration if the same block is attached to a spring of 500 newtons per meter? So the mass is constant here.

The only thing that changes is the spring constant and f. So we know that f is equal to 1 /

2<unk>i *<unk> of k / m. So we can write a ratio f2 /

f_sub_1 is 1 / 2 pi square<unk> k2 m. So we we're going to write the subscript for k since k changes. But m is

constant. So we don't need to write the subscript for m. So we can cancel the 2 pi and the mass. So we're going to get

this equation f_sub_2 / f_sub_1 is equal to the square t of k2 / k1. So as you increase the spring constant

the force will increase. So I mean not the force but the frequency will increase. So the

frequency should be higher than 15 htz. So f_sub_1 is 15 and that corresponds to a k constant of 200. K2 is 500 and we're

looking for f_sub_2. So if we divide 500 by 200, that's 2.5. And the square root of 2.5 is 1.581.

So f_sub_2 / 15 is that number. So to find f2, we need to multiply 1.581 by 15 and you should get a frequency of 23.7

hertz. A block under goes SHM with amplitude point4. That's simple harmonic motion.

If you see sho that's a simple harmonic oscillator. What is the total distance that it

travels in eight periods? Now keep this in mind the distance that it travels in one

period or one cycle is four times the amplitude. Let's say this is the equilibrium position.

This is the amplitude, the amount that it's fully stretched by. So

for to for this block to travel one complete cycle, it has to travel a towards the equilibrium position and

another distance of A and then back towards the equilibrium position and back towards where it started. So as you

can see the distance that it travels in one cycle or one period is 4 * a. So 4 *4

is 1.6 m. So in one period it's going to travel a distance of 1.6 m. In 8 periods it's going to be 1.6 * 8

which is about 12.8 m. So that's the total distance that the block will travel in a periods.

Now there's some other functions that can describe the motion of a mass spring system.

So let's say if we have a vertical spring and currently it's compressed and let's

say this is the equilibrium position. So it's capable of bouncing up and down. So therefore this is going to be equal

to the amplitude and this will be a Now as it bounced up and down, if you

plot this on a graph, let's say time is the x- axis

and the displacement is on the y- axis. So this is going to be the maximum displacement and here's negative a.

So according to this graph, it's starting at its compressed position. So that's at the top. If you graph it over

time, it's going to form a sinosoidal function. It's basically a sine wave. Now, because

it started at the top, this is the cosine function. If it started at the middle, it would be s.

Now if we draw the circle that we had earlier in this video, this is x and the radius of the circle

is a on the inside. This is the angle theta and this is the right triangle.

According to uh circa, if you've taken trigonometry at this point, you know that coine

of the angle theta is equal to the adjacent side / the hypotenuse. That's the cup part in SOA.

The adjacent side is the side that's very close to the angle. So that's X. The hypotenuse is across the box. That's

a. So cosine theta is X / A. So if you multiply both sides by A, you can get the value of X.

So therefore x is equal to a cossine theta. Now we know that linear displacement is

equal to velocity multiplied by time d= vt. Angular displacement is equal to omega

which is angular velocity multiplied by time. Now perhaps you have seen this equation

as you've studied rotational kinematics. If you haven't, you can look up my video that I've created on YouTube. You can

check that out and you'll see that equation. Omega, which is the angular velocity,

it's equal to 2<unk>i * the frequency. So therefore, we could say x is equal to a cosine

omega * t since theta equals omega * t. And then we can replace omega with 2 pi f. So x the

position function with respect to time is equal to the amplitude time cosine 2 pi f * t. So make sure that you write

this equation down. This is an equation that you need to know.

Now if you taken calculus you know that the derivative of the position function x gives you the instantaneous velocity.

So let's find the derivative. The derivative of cosine is equal to negative sign.

Now you have to keep the angle the same the 2 pi ft and then you have to differentiate the nested function or the

inside part of s which is 2 pi ft. The derivative of 2 pi ft is simply 2 pi f. t is the variable that you're

differentiating with respect to. The derivative of 5x is 5. The derivative of 9x is 9. The derivative of 8t is 8. So

the derivative of 2 pi ft is 2 pi f. By the way, for those of you who haven't taken calculus and who want to

understand how to find the derivative of a function to understand this step, you can find my video on derivatives in

YouTube. You just got to search for it. You may have to type in like power rule, product rule, quotient rule, but you

only need the power rule for this function. And also, you need to know how to find the derivative of trigonometric

functions as well, which you can also find out on YouTube. So, now let's continue.

So we need to replace a few things at this point. Let's replace f. We know that the

frequency is 1 / 2<unk>i time the<unk> of k / m.

In addition, earlier in this video, we mentioned that the amplitude is equal to the maximum velocity

times the square roo<unk> of m / k. And this came from the equation vax is equal to<unk> km