# Simple Harmonic Motion an Object Attached to a Spring

Simple Harmonic Motion .. an Object Attached to a Spring!!

Hi everybody .. i'm glad that i joined this forum so that i can help people and benefit at the same time :) .. i wanted to start with this problem right here ,, and hope that i get the hand from you :)

it's a question i've been working on for about a week and still no use ... and i really appreciate it if somebody just give some help soon ## Homework Statement

it's basically about two masses: m1 & m2 ,,, if m1 is the only mass attached to the end of a spring which is on a frictionless surface ,, and m2 is near m1 but not attached to anything..

The spring is disturbed from its equilibrium position because of a pushing force directed to the left (by a hand for example) with the two masses near each other ... if the force is released .. the spring will exert a restoring force (F.elastic) directed toward the equilibrium position (which is on the right of the masses) .. now the masses will go to the right near each other to a point where they will separate because m2 is not attached to m1 or to the spring ..

the questions are :
1- when will the two masses separate (m2 leave and separate from m1)
2- calculate the time needed for m1 (which is only left after m2 had left the system) to get back to its starting place on the left of the equilibrium position..

I'm sure the answer to the question won't be a number because there are no information enough to the get just a number in the solution ,, but the question is just right and there are no mistakes or lacks of data in it .. and i think the answer to the first question may consists of a proportion of the Period (T) but i'm not sure!!

## Homework Equations

NOTE ..

T = 2 $$\Pi$$ $$\sqrt{\frac{m}{k}}$$ or T = 2(pi)(sqrt(m/k))

HOOKE’S LAW: Fe = -kx
and other laws of motion and forces

Thanks in advance :D ,, and to anybody who would try to help me out :-)
i'll be glad to show me the way of solving :)

## The Attempt at a Solution

i couldn't get anywhere trying to solve this problem for almost a week !!

#### Attachments

• 11.8 KB Views: 510
Last edited:

Related Introductory Physics Homework Help News on Phys.org
ehild
Homework Helper

You have two time periodes, T and T' one with the two masses together, the other with mass m1 alone.
Draw the displacement from equilibrium of mass m1 as function of t/T. When is m1 at the farthest position on the right?

From here, it moves backwards, but it can not pull m2: it will move alone effected by the spring. What is its time period ?

ehild

anyway ,, i think i understood like 50% of what you have said :(

The thing is that the two masses (m1 & m2) .. starts accelerating from the left toward the equilibrium position where x=0 then the spring starts to decelerate and its speed will get slower .. from this point i think that m2 will complete from the equilibrium position with Vi (initial velocity) towards the right like a falling object with an initial horizental speed ,, but m1 as it crosses the equilibrium position ,the force Fe will pull it back and it's Velocity starts decreasing with an acceleration to the left .. So i think m1 & m2 will separate at the equilibrium position .. but still i'm not sure!!

about the time period (T) ,, i don't know it ,, that it wasn't mentioned in the question our teacher gave it to us .. fortunately he said that you don't need it and the answer will consists of a proportion of it .. !!! but i didn't get it and i'm not sure!!

you said to draw the displacement as a function of t/T .. Does it mean x=t/T ??

You have two time periodes, T and T' one with the two masses together, the other with mass m1 alone.
Draw the displacement from equilibrium of mass m1 as function of t/T. When is m1 at the farthest position on the right?

From here, it moves backwards, but it can not pull m2: it will move alone effected by the spring. What is its time period ?

ehild
I think you're mistaken.
You should draw the velocity as a function of displacement from equilibrium (You only need the maximum, to be honest)
The spring can both push and pull, but $$m_1$$ can only push and not pull on $$m_2$$

Find the point where their velocities start to differ, that's when they disconnect.

ok :) i did think like finding the point where they have different velocities .. i just don't know how to begin ,, and how should i draw the velocity as a function of displacement?!

thanx any way :D

ehild
Homework Helper

ok :) i did think like finding the point where they have different velocities .. i just don't know how to begin ,, and how should i draw the velocity as a function of displacement?!

thanx any way :D
I made an error, the velocities start to differ at the equilibrium length of the spring.

Till reaching the equilibrium point, it is simple harmonic motion. How much time elapses from maximum displacement to equilibrium?

ehild

anyway ,, i think i understood like 50% of what you have said :(

The thing is that the two masses (m1 & m2) .. starts accelerating from the left toward the equilibrium position where x=0 then the spring starts to decelerate and its speed will get slower .. from this point i think that m2 will complete from the equilibrium position with Vi (initial velocity) towards the right like a falling object with an initial horizental speed ,, but m1 as it crosses the equilibrium position ,the force Fe will pull it back and it's Velocity starts decreasing with an acceleration to the left .. So i think m1 & m2 will separate at the equilibrium position .. but still i'm not sure!!

about the time period (T) ,, i don't know it ,, that it wasn't mentioned in the question our teacher gave it to us .. fortunately he said that you don't need it and the answer will consists of a proportion of it .. !!! but i didn't get it and i'm not sure!!

you said to draw the displacement as a function of t/T .. Does it mean x=t/T ??
That's exactly right!
From the starting position up until $$x=0$$ the spring force pushes on $$m_1$$ which in turn pushes on $$m_2$$
They then reach the equilibrium position with some maximal velocity (Energy considerations should help you find this velocity without mucking about with the SHM equations, if you prefer it that way)
At this point, the spring force turns into a pulling force. It pulls on $$m_1$$, decreasing its velocity, but from this point on, $$m_2$$ experiences no force so it continues its motion to the right, without a change in velocity (Newton's first law).

As for how long this will take, how much of a period is it from one amplitude extreme to the equilibrium? Remember that one period is:
Left extreme - equilibrium - right extreme - equilibrium - left extreme

Now that you know what fraction of a period has passed, try and find the entire period with the two masses, $$T_{1+2}$$
You remember the SHM formulas, so $$T_{1+2}$$ should be easy to find!

The second section is a bit trickier though. Are they asking for the time it takes for it to return to the "left extreme" position, or the time it takes for it to return to the exact same position it started in?
The latter requires quite a few calculations (Are you familiar with the SHM equations of motion?), while the former is much more straight-forward.

guys to be honest ,, it's not a simple question at all !!

as you may think that it's just a question you may face and solve in a minutes!!
you know our teacher said that whoever solve it will get a full mark in all the exams without taking them ,, and since a week nothing happened :D

and about quarter the time period (T) ,it isn't told in the question ..

well , i think that the whole question is a mathematics trick ,, what i've done so far is to replace the k with -Fe/x and then Fe with m.a because (F=ma) and Fe is the only force applied on the mass and then to omit the m with the m ,, and with some math and maybe laws of motion the answer may reveal!!

it's one of the questions i'll never be able to solve :D
anyway thnx alot :) i appreciate your help.

guys to be honest ,, it's not a simple question at all !!

as you may think that it's just a question you may face and solve in a minutes!!
you know our teacher said that whoever solve it will get a full mark in all the exams without taking them ,, and since a week nothing happened :D

and about quarter the time period (T) ,it isn't told in the question ..

well , i think that the whole question is a mathematics trick ,, what i've done so far is to replace the k with -Fe/x and then Fe with m.a because (F=ma) and Fe is the only force applied on the mass and then to omit the m with the m ,, and with some math and maybe laws of motion the answer may reveal!!

it's one of the questions i'll never be able to solve :D
anyway thnx alot :) i appreciate your help.
You've posted the formula for the period yourself!

$$T=\frac{2\pi}{\omega}$$
$$\omega=\sqrt{\frac{k}{m}}$$

Where $$m$$ is the mass that the spring has to push/pull.

thanks a lot RoyalCat :) you just got right :D

but isn't the left extreme the same as the point the whole system started from when the force of a hand was released? .. remember that the length of an oscillatin of a spring isn't determined by its mass ,, but by the place where you pushed it and then released the force so it will oscillate and come back to that exact location and turn its direction in it .. what matters when you change the mass is the period (T) and that's the only thing that's determined by the mass ,, not the amplitude!!

about the Simple harmo..equations .. if you mean the derivatives ,, no we didn't take them .. we only studied the equations above in the main post .. and they should work ,, everything that has a connection to derivatives in physics we didn't take that ,, we've just studied deriving in math so.. :S

thanks for the great help :D

You've posted the formula for the period yourself!

$$T=\frac{2\pi}{\omega}$$
$$\omega=\sqrt{\frac{k}{m}}$$

Where $$m$$ is the mass that the spring has to push/pull.
alright :) but what does omega represent?!

You've posted the formula for the period yourself!

$$T=\frac{2\pi}{\omega}$$
$$\omega=\sqrt{\frac{k}{m}}$$

Where $$m$$ is the mass that the spring has to push/pull.
again and i'm sorry ..

if omega = sqrt(k/m) then T= (2(pi))/(sqrt(k/m))
but we studied that T = 2(pi)(sqrt(k/m))

so the sqrt(k/m) is above not at the bottom .. i'm confused NOW :S

ehild
Homework Helper

You don't remember well. You wrote in your original post that
T = 2(pi)(sqrt(m/k)). This is correct.

ehild

thanks a lot RoyalCat :) you just got right :D

but isn't the left extreme the same as the point the whole system started from when the force of a hand was released? .. remember that the length of an oscillatin of a spring isn't determined by its mass ,, but by the place where you pushed it and then released the force so it will oscillate and come back to that exact location and turn its direction in it .. what matters when you change the mass is the period (T) and that's the only thing that's determined by the mass ,, not the amplitude!!

about the Simple harmo..equations .. if you mean the derivatives ,, no we didn't take them .. we only studied the equations above in the main post .. and they should work ,, everything that has a connection to derivatives in physics we didn't take that ,, we've just studied deriving in math so.. :S

thanks for the great help :D
Ah, so that's why this is an extra credit question. It is very hard to study SHM without understanding the key equation behind it.

So let's try and understand just what that omega is!

Let's first look at the simpler case of a mass, $$m$$ attached to a horizontal spring with spring constant $$k$$
What forces act on it? Only the force of the spring. That force is proportional to the displacement from equilibrium, $$x$$

$$\vec F_{spring}=-k\vec x$$

I've used vector notation to indicate that the force of the spring is always in the direction opposite the displacement. So if I pull the spring right (Let's define right as the positive $$x$$ direction) the mass will feel a force to the left, opposite the direction of the displacement.

And if I push the spring left, the mass will feel a force to the right, opposite the direction of the displacement.

We now write out Newton's Second Law for the mass with the spring attached.

$$\Sigma \vec F = m\vec a$$

Remembering that the only force is the spring force, we find:

$$ma = -kx$$

Rearranging we get:

$$a=-\frac{k}{m}x$$

We now define the following constant:

$$\omega ^2 \equiv \frac{k}{m}$$

Which brings us to the defining equation of SHM in mechanics:

$$a=-\omega^2 x$$

Now for the calculus part...
I'll try and go slowly, so feel free to ask any questions that arise.
Acceleration is the second derivative of position with respect to time.
It is the rate of change of the velocity with respect to time, and velocity itself is the rate of change of position with respect to time (This is basic kinematics, and just how these quantities, velocity and acceleration, are defined).

Mathematically:
$$a=\frac{dv}{dt}=\frac{d^2 x}{dt^2}$$

Us physicists have a special notation for time derivatives. The first time derivative of position is noted $$\dot x$$ and the second time derivative of position is noted $$\ddot x$$

Using this special notation we rewrite for some insight:

$$\ddot x = -\omega ^2 x$$

This equation is called a differential equation.
Don't let the name scare you, it is a fairly simple one.

Just like a regular equation with an unknown, for instance:

$$5-x=3$$

asks you "What number $$x$$ when subtracted from 5 gives you 3?"

The above differential equation asks you, "What function of time, $$x$$, when differentiated twice with respect to time, gives itself times a negative constant?"

By the end of the year, you should be able to answer that question in a heartbeat!

I don't know if you've reached the trigonometric functions in your calculus class yet, so just take what I'm saying as fact for now. You'll eventually prove the following relations:

$$\frac{d(\cos{(ax)})}{dx}=-a\sin{(ax)}$$

$$\frac{d(\sin{(ax)})}{dx}=a\cos{(ax)}$$

So let's examine a function of time, $$A\cos{(\omega t+\phi)}$$ where $$A$$ has dimensions of length, $$\omega$$ has dimensions of 1/time and $$\phi$$ is a pure number.

Taking its second derivative we arrive at:

$$-\omega^2 \cdot A\cos{(\omega t+\phi)}$$[/tex]

We've found a function whose second derivative is itself times a negative constant! This holds for any cosinusoidal or sinusoidal function!

So now we've found a general form for the displacement from equilibrium as a function of time of a simple harmonic oscillator.

$$x(t)=A\cos{(\omega t + \phi)}$$

Now we ask ourselves, how long does it take for the mass to reach the same spot (What is the period?)

To find the period, $$T$$, we need to solve the following equation:

$$x(t)=x(t+T)$$

Some trig later gives us the result you had given to you, $$T=\frac{2\pi}{\omega}$$

Taking the first derivative with respect to time, we find the velocity as a function of time:

$$v(t)=-\omega A\sin{(\omega t + \phi)}$$

We now have the tools required to solve both the second and the first parts of the question.

Ah, so that's why this is an extra credit question. It is very hard to study SHM without understanding the key equation behind it.

So let's try and understand just what that omega is!

Let's first look at the simpler case of a mass, $$m$$ attached to a horizontal spring with spring constant $$k$$
What forces act on it? Only the force of the spring. That force is proportional to the displacement from equilibrium, $$x$$

$$\vec F_{spring}=-k\vec x$$

I've used vector notation to indicate that the force of the spring is always in the direction opposite the displacement. So if I pull the spring right (Let's define right as the positive $$x$$ direction) the mass will feel a force to the left, opposite the direction of the displacement.

And if I push the spring left, the mass will feel a force to the right, opposite the direction of the displacement.

We now write out Newton's Second Law for the mass with the spring attached.

$$\Sigma \vec F = m\vec a$$

Remembering that the only force is the spring force, we find:

$$ma = -kx$$

Rearranging we get:

$$a=-\frac{k}{m}x$$

We now define the following constant:

$$\omega ^2 \equiv \frac{k}{m}$$

Which brings us to the defining equation of SHM in mechanics:

$$a=-\omega^2 x$$

Now for the calculus part...
I'll try and go slowly, so feel free to ask any questions that arise.
Acceleration is the second derivative of position with respect to time.
It is the rate of change of the velocity with respect to time, and velocity itself is the rate of change of position with respect to time (This is basic kinematics, and just how these quantities, velocity and acceleration, are defined).

Mathematically:
$$a=\frac{dv}{dt}=\frac{d^2 x}{dt^2}$$

Us physicists have a special notation for time derivatives. The first time derivative of position is noted $$\dot x$$ and the second time derivative of position is noted $$\ddot x$$

Using this special notation we rewrite for some insight:

$$\ddot x = -\omega ^2 x$$

This equation is called a differential equation.
Don't let the name scare you, it is a fairly simple one.

Just like a regular equation with an unknown, for instance:

$$5-x=3$$

asks you "What number $$x$$ when subtracted from 5 gives you 3?"

The above differential equation asks you, "What function of time, $$x$$, when differentiated twice with respect to time, gives itself times a negative constant?"

By the end of the year, you should be able to answer that question in a heartbeat!

I don't know if you've reached the trigonometric functions in your calculus class yet, so just take what I'm saying as fact for now. You'll eventually prove the following relations:

$$\frac{d(\cos{(ax)})}{dx}=-a\sin{(ax)}$$

$$\frac{d(\sin{(ax)})}{dx}=a\cos{(ax)}$$

So let's examine a function of time, $$A\cos{(\omega t+\phi)}$$ where $$A$$ has dimensions of length, $$\omega$$ has dimensions of 1/time and $$\phi$$ is a pure number.

Taking its second derivative we arrive at:

$$-\omega^2 \cdot A\cos{(\omega t+\phi)}$$[/tex]

We've found a function whose second derivative is itself times a negative constant! This holds for any cosinusoidal or sinusoidal function!

So now we've found a general form for the displacement from equilibrium as a function of time of a simple harmonic oscillator.

$$x(t)=A\cos{(\omega t + \phi)}$$

Now we ask ourselves, how long does it take for the mass to reach the same spot (What is the period?)

To find the period, $$T$$, we need to solve the following equation:

$$x(t)=x(t+T)$$

Some trig later gives us the result you had given to you, $$T=\frac{2\pi}{\omega}$$

Taking the first derivative with respect to time, we find the velocity as a function of time:

$$v(t)=-\omega A\sin{(\omega t + \phi)}$$

We now have the tools required to solve both the second and the first parts of the question.

your explanation was really more than awesome :) ,, but there's only one thing i didn't really get it, that's when you said : "So let's examine a function of time, $$A\cos{(\omega t+\phi)}$$" ,, is it because the drawing of cos(x) function is similar to the oscillation of a spring ,, if so then i guess the A represent Amplitude ,, but what's [(omega)(t)+(phi)]? is phi the golden ratio?

and how did you get these equations :

$$x(t)=x(t+T)$$
$$T=\frac{2\pi}{\omega}$$

and my final problem :D .. as i mentioned that there's two different Periods (T), one with the two masses for quarter the oscillation and one with one mass for three quarters of one complete oscillation ,, that is m1 ...
how would i be able to find T ,, if there are two different Ts or periods!? or will there be an average period ?!

Everything else was as clear as a sun :)

thank you for your help ,, you are the best :D
i guess my teacher wouldn't have helped me this way your explanation was really more than awesome :) ,, but there's only one thing i didn't really get it, that's when you said : "So let's examine a function of time, $$A\cos{(\omega t+\phi)}$$" ,, is it because the drawing of cos(x) function is similar to the oscillation of a spring ,, if so then i guess the A represent Amplitude ,, but what's [(omega)(t)+(phi)]? is phi the golden ratio?

and how did you get these equations :

$$x(t)=x(t+T)$$
$$T=\frac{2\pi}{\omega}$$

and my final problem :D .. as i mentioned that there's two different Periods (T), one with the two masses for quarter the oscillation and one with one mass for three quarters of one complete oscillation ,, that is m1 ...
how would i be able to find T ,, if there are two different Ts or periods!? or will there be an average period ?!

Everything else was as clear as a sun :)

thank you for your help ,, you are the best :D
i guess my teacher wouldn't have helped me this way That was an educated guess, since I know that when I take the second derivative of a trig function, I get the function itself times a negative constant.
And yes, since the trig functions are one of the few periodic functions we know, that too could have been as good a reason as any to use them as our trial solution.

There are more rigorous ways to get to the solution, but we just guessed it to be a sinusoidal or cosinusoidal function.

$$\phi$$ is what's called a phase angle. It is determined by our initial conditions and is nothing more than an angle.

To make the point more clear, $$A\cos{(\omega t + \phi)}$$ is just our general solution. $$x(t)$$ is of the FORM $$A\cos{(\omega t + \phi)}$$, but what is the amplitude of the motion ($$A$$)? and what is the phase angle?

For that, we need to have two additional pieces of information on the motion of the object, for instance, its position and velocity at time $$t=0$$

In our case, the initial displacement $$x(t=0) = D$$ (Some maximal displacement we are not told, it is probably irrelevant)
and the initial velocity $$v(t=0) = 0$$

Plugging that into our equations of motion yields:

$$A\cos{(\omega \cdot 0 + \phi)}=D$$

$$A\cos{\phi}=D$$

On to the velocity:

$$-\omega A\sin{(\omega \cdot 0 + \phi)}=0$$

$$\sin{\phi}=0$$

Solving those two equations tells us the EXACT form of $$x(t)$$

$$x(t)=D\cos{(\omega t)}$$

In this specific case, $$\phi$$ is 0. But you can see that if the velocity at time $$t=0$$ were any different, we'd have a specific phase angle $$\phi$$.

$$T$$ is the length of time that satisfies, for any point in time, $$t$$, that once it has passed, the object returns to the exact same spot with the exact same velocity and acceleration.
It's how we define the period.

And how do we phrase that mathematically?

$$x(t)=x(t+T)$$
and
$$v(t)=v(t+T)$$

Let's use our function $$x(t)=D\cos{(\omega t)}$$

Plugging it into the "period defining equation" yields:

$$D\cos{(\omega t)}=D\cos{(\omega (t+T))}$$

$$\cos{(\omega t)}=\cos{(\omega (t+T))}$$

And from the velocity:

$$\sin{(\omega t)}=\sin{(\omega (t+T))}$$

Solving those trig equation is straightforward.
The only way both the sine and cosine functions return to the same value, is if we move through $$2\pi$$ radians. And the value of $$T$$ required for that is $$T=\frac{2\pi}{\omega}$$

You know exactly how to find the total time! You just phrased it yourself. :) You know how to find both of the expressions on the right-hand-side of this equation:

$$t_{total}=\tfrac{1}{4}T_{1+2}+\tfrac{3}{4}T_1$$

On a side note, please avoid quoting lengthy posts, there's no need to have my posts twice on the page. :p

Thanks Alot :D
you really helped me out and solved my question ,, i appreciate it :)

about that side note :P ,, i only did it to get the equations u mentioned :D

Happy Holidays :)