Understanding the Harmonic Motion of Two Masses Connected by a Spring

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

i)
$x_{cm} = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}$
At any time t , $x_{cm} = v_0t$ and
$x_1= v_0t - A(1-\cos{ωt})$
From the above two ,

we get $x_2 = v_0t + \dfrac{m_1}{m_2}A(1-\cos{wt})$

I'm not clear what to do in part (ii) .

I might have to use Energy conservation between times t=0 and time when the spring is in maximum elongation , but then what is the time when this happens and what is the maximum elongation of spring . Will it be the same amount by which it is compressed i.e L₀
?

 

[haruspex]

what is the maximum elongation of spring . Will it be the same amount by which it is compressed i.e L0

To answer that, consider energy totals.

 

[Jahnavi]

To answer that, consider energy totals.

I get $A = \frac{m_2 L_0}{2(m_1+m_2)}$

 

[haruspex]

I get $A = \frac{m_2 L_0}{2(m_1+m_2)}$

It cannot be asymmetric in m1, m2. Please show your working.

Edit: I retract that. I did not register that the definition of A is asymmetric in m1, m2.

 

[Jahnavi]

It cannot be asymmetric in m1, m2.

x₁(t) has m₂ and x₂(t) has m₁ in the numerator in their second terms respectively .

I rechecked my work .There should not be a 2 in the denominator .

Now the answer matches with the answer key

If you think $A=\frac{m_2L_0}{m_1+m_2}$ is incorrect please let me know .

 

[haruspex]

x₁(t) has m₂ and x₂(t) has m₁ in the numerator in their second terms respectively .

I rechecked my work .There should not be a 2 in the denominator .

Now the answer matches with the answer key

If you think $A=\frac{m_2L_0}{m_1+m_2}$ is incorrect please let me know .

Sorry, you are right. It is asymmetric because of the way A is defined.

 

[Jahnavi]

Thank you .

Is the motion of m₁ and m₂ SHM with respect to the Center of Mass ?

 

[haruspex]

Thank you .

Is the motion of m₁ and m₂ SHM with respect to the Center of Mass ?

Certainly.

 

[Jahnavi]

Certainly.

SHM is of the form kcos(ωt+Φ) , but in this question it is k(1-cosωt) ?

Are the two equivalent ?

 

[haruspex]

SHM is of the form kcos(ωt+Φ) , but in this question it is k(1-cosωt) ?

Are the two equivalent ?

It is still SHM, just around a different equilibrium point.
Perhaps I misunderstood your question. Did you mean, it is SHM with the common mass centre as equilibrium point? If so, no.

 

[Jahnavi]

Are m1 and m2 performing SHM as seen from Center of Mass frame ?

If so , then x₁(t) and x₂(t) should be equal to x_cm + kcos(ωt+Φ) ?

This is not happening in this problem .

Instead they are written as x_cm + k(1-cos(ωt) .

 

[haruspex]

Are m1 and m2 performing SHM as seen from Center of Mass frame ?

If so , then x₁(t) and x₂(t) should be equal to x_cm + kcos(ωt+Φ) ?

This is not happening in this problem .

Instead they are written as x_cm + k(1-cos(ωt) .

Your equation for SHM is unnecessarily restrictive. Let me quote Wikipedia:

The motion of a particle moving along a straight line with an acceleration which is always towards a fixed point on the line and whose magnitude is proportional to the distance from the fixed point is called simple harmonic motion [SHM]​

In the centre of mass frame, we can choose a point at offset k as that fixed point. It does not need to be the origin of the frame.

 

[Jahnavi]

In the centre of mass frame, we can choose a point at offset k as that fixed point. It does not need to be the origin of the frame.

Sorry I don't understand . Please explain in the context of above setup .

Is x = A(1-cosωt) an SHM ?

 

[haruspex]

Sorry I don't understand . Please explain in the context of above setup .

Is x = A(1-cosωt) an SHM ?

Yes. There is a fixed point, x=A, such that the acceleration towards it is proportional to the distance from it.

 

[Jahnavi]

Yes. There is a fixed point, x=A, such that the acceleration towards it is proportional to the distance from it.

So , even though the frame (CM frame) is moving , the acceleration is proportional to a fixed point .

Where is that fixed point in this problem ?

 

[haruspex]

the acceleration is proportional to a fixed point .

A point that is fixed within the frame.
In the ground frame it is not SHM. in the common mass centre frame there is a fixed point (k) satisfying the conditions to make it SHM.

 

[Jahnavi]

A point that is fixed within the frame.
In the ground frame it is not SHM. in the common mass centre frame there is a fixed point (k) satisfying the conditions to make it SHM.

Is the fixed point in the frame necessarily the point of equilibrium for the SHM in that frame ?

In this problem , for m2 , is the fixed point a distance $\frac{m_1L_0}{m_1+m_2}$ from the CM (origin) ?

Similarly , for m1 , is the fixed point a distance $\frac{m_2L_0}{m_1+m_2}$ from the CM (origin) ?

 

[Jahnavi]

$x_2 = v_0t + \dfrac{m_1}{m_2}A(1-\cos{wt})$

So , in the CM frame the displacement of m₂ from the fixed point is basically $-cosωt$ ?

$-cosωt = cos(ωt+π)$

With respect to the fixed point in the CM frame, the displacement of m₂ turns out to be x=kcos(ωt+π) which is indeed an SHM .

Is my understanding correct ?

 

[haruspex]

Is the fixed point in the frame necessarily the point of equilibrium for the SHM in that frame ?

In this problem , for m2 , is the fixed point a distance $\frac{m_1L_0}{m_1+m_2}$ from the CM (origin) ?

Similarly , for m1 , is the fixed point a distance $\frac{m_2L_0}{m_1+m_2}$ from the CM (origin) ?

$x_2 = v_0t + \dfrac{m_1}{m_2}A(1-\cos{wt})$

So , in the CM frame the displacement of m₂ from the fixed point is basically $-cosωt$ ?

$-cosωt = cos(ωt+π)$

With respect to the fixed point in the CM frame, the displacement of m₂ turns out to be x=kcos(ωt+π) which is indeed an SHM .

Is my understanding correct ?

Yes to all the above.