How Does the Period of Oscillation Change When a Block Detaches from the Wall?

[Vibhor]

Homework Statement

A system consists of two blocks, each of mass M, connected by a spring of force constant k. The system is initially shoved against a wall so that the spring is compressed a distance D from its original uncompressed length. The floor is frictionless. The system is now released with no initial velocity.

Determine the period of oscillation for the system when the left block is no longer in contact with the wall.

Homework Equations

The Attempt at a Solution

I am finding it difficult to understand the motion of the system after the left block leaves the wall.

After the left block leaves the wall ,the CM moves with constant velocity since there would be no external force ,but how will the blocks move ?

I know how to deal with the problem where there is no left block and the left end of the spring is attached to the wall .But this case is different .

Please help me in finding the period of oscillation

 

[rude man]

OK, first decide on what frequency you are looking for. Realize that both blocks have enough energy to move out towards infinity so looking for the displacement of the blocks individually might not be a good idea. It's also very messy math.

So how about asking for the frequency of the distance variation of the blocks from each other with time?
What would be the F=ma equations for the two blocks from the instant the left block leaves the wall?

 

[Vibhor]

OK, first decide on what frequency you are looking for. Realize that both blocks have enough energy to move out towards infinity so looking for the displacement of the blocks individually might not be a good idea. It's also very messy math.

So how about asking for the frequency of the distance variation of the blocks from each other with time?

.Are there different type of frequencies involved in an oscillation ? How do I know which frequency is being asked in the question ?

What would be the F=ma equations for the two blocks from the instant the left block leaves the wall?

Equal and opposite forces will act on the two blocks.If F=kx on left block ,then F=-kx on right block .

 

[SammyS]

Homework Statement

A system consists of two blocks, each of mass M, connected by a spring of force constant k. The system is initially shoved against a wall so that the spring is compressed a distance D from its original uncompressed length. The floor is frictionless. The system is now released with no initial velocity.

Determine the period of oscillation for the system when the left block is no longer in contact with the wall.

Homework Equations

The Attempt at a Solution

I am finding it difficult to understand the motion of the system after the left block leaves the wall.

After the left block leaves the wall ,the CM moves with constant velocity since there would be no external force ,but how will the blocks move ?

I know how to deal with the problem where there is no left block and the left end of the spring is attached to the wall .But this case is different .

Please help me in finding the period of oscillation

You are correct to conclude that "After the left block leaves the wall ,the CM moves with constant velocity since there would be no external force."

Subtract that constant velocity from the system as a whole, then solve that problem. (Add that velocity to everything for the overall solution, if you need it.)

What can you say about the motion of each block relative to the Center of Mass?

 

[Vibhor]

What can you say about the motion of each block relative to the Center of Mass?

From the ground frame ,accelerations will be equal in magnitude and opposite in direction whereas the speeds will be different .The velocities will be such that the CM moves with constant velocity.

From the CM frame , accelerations as well as velocity will be equal in magnitude and opposite in direction .

 

[rude man]

.Are there different type of frequencies involved in an oscillation ? How do I know which frequency is being asked in the question ?



Equal and opposite forces will act on the two blocks.If F=kx on left block ,then F=-kx on right block .

No, there is only one frequency, but the time domain solution for both masses also icludes a term proportional to time.

I suggested solving for the distance between the two masses since that is a simple sine motion.

yes, as you say, equal and opposite forces act on the masses.The force is zero when the spring is relaxed, of course, and is kx and -kx for the left and right masses respectively, again as you say, with x being the distance between the masses minus the relaxed length of the spring.

 

[Vibhor]

How should I proceed ?

 

[rude man]

How should I proceed ?

Write equations for each of the two masses: f = ma. Call the length of the relaxed spring L. Call the displacement of the left mass x1 and that of the right-hand one x2. So if the spring is relaxed, x2 - x1 = L. What do these two equations look like?

 

[Vibhor]

For m1 , ma1 = k(x2-x1-L)

For m2 , ma2 = -k(x2-x1-L)

 

[BiGyElLoWhAt]

Hmm... it seems like this is being made harder than it needs to be.

If you look at the time period between when it was initially compressed and when the left block left the wall, what can you say about certain aspects of the oscillation?

Now what can you say will remain the same despite the initial conditions (I'm thing of one unknown that's in terms of 2 knowns that you have)

From answering these 2 questions you should be able to model the motion of the 2 blocks with respect to the center of mass of the spring.

Then you just need to add in the constant velocity from the motion of the system as a whole.

 

[AlephZero]

If you look at the time period between when it was initially compressed and when the left block left the wall, what can you say about certain aspects of the oscillation?

Now what can you say will remain the same despite the initial conditions (I'm thing of one unknown that's in terms of 2 knowns that you have)

That is a dangerous way to try to answer the question. The vibration frequency with one block in contact with the wall is not the same as when both blocks are free to move. Or looking at it a different way, while one block is in contact with the wall, the center of mass of both blocks in not moving at constant velocity, it is accelerating.

You could get the right answer that way if you are smart enough, but not if you are just learning dynamics IMO.

Vibhor is doing OK with the equations of motion for m1 and m2.

 

[rude man]

For m1 , ma1 = k(x2-x1-L)

For m2 , ma2 = -k(x2-x1-L)

Very good! But now, what are a1 and a2 in terms of x1 and x2 respectively?

 

[BiGyElLoWhAt]

while one block is in contact with the wall, the center of mass of both blocks in not moving at constant velocity, it is accelerating.

I never said it was. But seeing as how the net force on the COM is 0 after the left block leaves the wall, it will move at a constant velocity from then on.

And I agree that the currently used method will work, I just thought you would get a "nicer" answer doing it using the concepts I mentioned earlier.

 

[rude man]

@other posters: once the left-hand mass leaves the wall it never returns to the wall. Both masses and the difference in distance between them all oscillate with the same frequency. As long as the left-hand mass rests against the wall there are no oscillations. The left-hand mass moves forward as soon as the spring is relaxed for the first time (after the initial compression).

P.S. OP, I would not spend a lot of time worrying about what the center of mass is doing. Solve your two equations and believe the math!

 

[SammyS]

All that's asked for is the oscillation frequency of the system after the left hand block loses contact with the wall.

Taking the Center of Mass into consideration can make the solution quite simple, provided that the OP is familiar with the frequency of oscillation for a simple mass/spring system.


What is the effective spring constant for the spring which is produced by cutting a linear spring (with spring constant, k) in half ?

 

[AlephZero]

As long as the left-hand mass rests against the wall there are no oscillations.

While the left hand mass is in contact with the wall, the motion is the same as if the left hand end of the spring is fixed to the wall. The other mass starts doing a simple harmonic motion oscillation at one frequency, and then switches to a different frequency when the left hand mass leaves the wall.

I assumed that was what BiGyElLoWhAt meant.

If you are familiar with this sort of problem you could just write down the correct answers for both frequencies, but if the OP could do that he/she wouldn't have started the thread!

 

[rude man]

While the left hand mass is in contact with the wall, the motion is the same as if the left hand end of the spring is fixed to the wall. The other mass starts doing a simple harmonic motion oscillation at one frequency, and then switches to a different frequency when the left hand mass leaves the wall.

There are no oscillations until the mass leaves the wall. As soon as the spring reaches relaxation the first time the left mass leaves the wall. Then they both oscillate. There is only one frequency.

 

[BiGyElLoWhAt]

rude man

There is 1/2 an oscillation at 1 frequency, but yes, no complete oscillations until the whole system begins its translational motion.

 

[BiGyElLoWhAt]

That's almost like saying that a system of a spring attached to a wall at one end and a block on the other doesn't undergo oscillatory motion if I compress it, release it, and catch it at t = T/2

Yea it doesn't complete an oscillation, but it's still oscillatory motion.

 

[rude man]

There is 1/2 an oscillation at 1 frequency, but yes, no complete oscillations until the whole system begins its translational motion.

It's not an oscillation. It's a monotonically increasing motion in the direction of original motion until the spring is relaxed and the left-hand mass leaves the wall.

Let's try to help the OP.

 

[BiGyElLoWhAt]

Let's try to help the OP.

Agreed

 

[Vibhor]

Very good! But now, what are a1 and a2 in terms of x1 and x2 respectively?

a1 = (k/m)(x2-x1-L)

a2 = -(k/m)(x2-x1-L)

What should I do next ?

Taking the Center of Mass into consideration can make the solution quite simple, provided that the OP is familiar with the frequency of oscillation for a simple mass/spring system.

For a simple mass/spring system time period is 2π√(m/k) .

What is the effective spring constant for the spring which is produced by cutting a linear spring (with spring constant, k) in half ?

2k is the spring constant for half the spring .

Are you suggesting that the time period is 2π√(m/2k) ?

Doesn't the motion of CM makes this problem different from the case when CM is stationary and the two blocks are oscillating ?

 

[SammyS]

In response to my previous post, you wrote:

2k is the spring constant for half the spring .

Are you suggesting that the time period is 2π√(m/2k) ?

Doesn't the motion of CM makes this problem different from the case when CM is stationary and the two blocks are oscillating ?

Yes, 2k is the spring constant.

The CM moves with constant velocity (Acceleration is zero.) after system breaks contact with the wall.

That is an inertial frame.

So yes, I'm suggesting that period for the oscillations -- with proper parentheses added.

2π√(m/(2k))

 

[Vibhor]

That is an inertial frame.

So yes, I'm suggesting that period for the oscillations -- with proper parentheses added.

2π√(m/(2k))

Do you mean that the period of oscillation in every inertial frame is same i.e in the CM frame as well as ground frame ?

The problems I have done so far consider oscillations around a fixed point about which the mass oscillates ,but here we have a moving point i.e CM ?

 

[rude man]

a1 = (k/m)(x2-x1-L)

a2 = -(k/m)(x2-x1-L)

What should I do next ?

You haven't expressed a1 and a2 in terms of x1 and x2. Hint: acceleration is the second time derivative of displacement ...

 

[Vibhor]

You haven't expressed a1 and a2 in terms of x1 and x2. Hint: acceleration is the second time derivative of displacement ...

d²x₁/dt² = (k/m)(x2-x1-L)

d²x₂/dt² = -(k/m)(x2-x1-L)

What next ?

 

[SammyS]

Do you mean that the period of oscillation in every inertial frame is same i.e in the CM frame as well as ground frame ?

The problems I have done so far consider oscillations around a fixed point about which the mass oscillates ,but here we have a moving point i.e CM ?

This does bring into question "What precisely do we mean by an oscillation?" -- or "What is a period of this motion?"

In this case, it's only in the frame of the CM, that there is truly periodic motion.

 

[Vibhor]

This does bring into question "What precisely do we mean by an oscillation?" -- or "What is a period of this motion?"

In this case, it's only in the frame of the CM, that there is truly periodic motion.

This is precisely what I was thinking .

Is it correct to think that the blocks are oscillating only about CM and do not oscillate in the ground frame ?

Was that implicit in the question ?

 

[rude man]

This is precisely what I was thinking .

Is it correct to think that the blocks are oscillating only about CM and do not oscillate in the ground frame ?

Was that implicit in the question ?

No, there is oscillation for each mass with respect to the observer, it's just that there is also constant velocity on top of the mass oscillations. That's why i suggested you focus on the distance between the two masses, which has only a sine oscillatory term in it.

In other words, each mass displacement looks like at + b sin(wt).

 

[Vibhor]

Please look at post#26 . What should I do next ?

 

[rude man]

d²x₁/dt² = (k/m)(x2-x1-L)

d²x₂/dt² = -(k/m)(x2-x1-L)

What next ?

Yay team! OK now, you need to subtract one equation from the other to form a single diff. eq. in (x2 - x1). Think carefully about your initial conditions. They're very simple for x1, more elaborate for x2 since x2 starts at x2=L and it also has an initial velocity when x2 = L.

In other words, you're forming a new variable x = x2 - x1. x is now the distance between the two masses.

PS you don't need to compute the initial velocity of the right-hand mass. Just call it x-dot-0 or whatever.

 

[Vibhor]

Yay team! OK now, you need to subtract one equation from the other to form a single diff. eq. in (x2 - x1). Think carefully about your initial conditions. They're very simple for x1, more elaborate for x2 since x2 starts at x2=L and it also has an initial velocity when x2 = L.

In other words, you're forming a new variable x = x2 - x1. x is now the distance between the two masses.

d²(x₂-x₁)/dt² = -(2k/m)(x2-x1-L)

If x=x2-x1 ,then

d²x/dt² = -(2k/m)(x-L)

or, d²x/dt² + (2k/m)x = 2kL/m

At t=0 (i.e when left block leaves the wall) , x = L , but how do I know what is dx/dt at t=0 ?

What next ?

 

[rude man]

d²(x₂-x₁)/dt² = -(2k/m)(x2-x1-L)

If x=x2-x1 ,then

d²x/dt² = -(2k/m)(x-L)

At t=0 (i.e when left block leaves the wall) , x = L , but how do I know what is dx/dt at t=0 ?

What next ?

You don't have to know the actual number of dx/dt(t=0). Just call it something (like x-dot-0).
Next is solving your diff. eq. Have you done diff eq's? If not you should look at the diff. eq. for simple harmonic motion and the frequency should pop right out for you. Your big hint is the term 2k/M. Remember what the frequency was if the diff. eq. is d²x/dt² + (k/M)x = constant? That would be a single mass dangling on a single spring with some initial deflection of the mass away from its rest position. That's all you really need to answer the problem.

 

[Vibhor]

Thank you rude man for the guidance.

You don't have to know the actual number of dx/dt(t=0). Just call it something (like x-dot-0).

We can find dx/dt(t=0) .The right block is compressed by D initially .Applying the conservation of energy ,

(1/2)mv₂² = (1/2)kD²

dx/dt at t=0 is equal to the speed of the right block when the left block leaves the wall.

dx/dt(at t=0) = D√(k/m)

So now we have the initial conditions at t=0 -> x=L ,dx/dt = D√(k/m) , d²x/dt² = 0

Do you think this is correct ?

 

[SammyS]

...

In this case, it's only in the frame of the CM, that there is truly periodic motion.

I should have said,

The only inertial frame for which the motion is truly periodic, is the Center of Mass frame.


A frame of reference fixed to either mass is a non-inertial frame. The problem can certainly be solved in such a reference frame, but beware of the possibility of fictitious forces.