Maximum extension in a system with two blocks separated by a spring

[Hamiltonian]

Homework Statement

A block of mass m is connected to another block of mass M by a massless spring of spring constant k. The blocks are kept on a smooth horizontal plane and are at rest. The spring is unstretched when a constant force F starts acting on the block of mass M to pull it. Find the maximum extension of the spring.

Relevant Equations

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the acceleration of the center of mass is $a_{cm} = F/(M+m)$
I considered the forces on the block of mass m(when the system is at maximum extension) I got the equation $$kx - \frac {mF}{(M+m)} = ma_{cm}$$
and from that I got the value of the maximum extension $$x = \frac {2mF} {k(M+m)}$$ which is the correct answer.
But when I considered the forces on the block of mass M(again when the system is at maximum extension) I get the equation $$F - kx - \frac {MF}{M+m} = Ma_{cm}$$
from which I got $$ x = \frac {F(m - M)}{k(m+M)}$$
(I have written these equations from the non-inertial frame attached to the system moving with the acceleration $a_{cm}$ both the blocks move with this acceleration only when there is maximum extension in the spring.)

the answer I got when considering forces on $M$ does not yield the same answer for the extension when I consider the forces on $m$.
I don't understand what I am doing wrong here(maybe this entire method is wrong as the correct answer is obtained by considering the system in the center of mass frame and then applying the work-energy theorem but if that's the case then I don't see what mistake I am making when I use Newtons laws the good old way)

 

[PeroK]

I think you might have got the right answer for the wrong reasons. The maximum extension is not when the acceleration of the blocks equals the acceleration of the CoM, but when the blocks come to rest in the CoM frame.

 

[Hamiltonian]

I think you might have got the right answer for the wrong reasons. The maximum extension is not when the acceleration of the blocks equals the acceleration of the CoM, but when the blocks come to rest in the CoM frame.

right after the instant when both blocks acquire the same velocity(i.e when there is maximum extension in the spring) won't both blocks from then on also have the same acceleration?

 

[pasmith]

There is an alternative method which uses the laboratory frame.

If the initial length of the spring is $d$, then the extension is $s = x_M - x_m - d$ where $x_M$ is the position of mass $M$, etc.

By considering the forces on each mass in an inertial frame, you can obtain $$\begin{align*} M\ddot x_M &= F - ks \\ m\ddot x_m &= ks \end{align*}$$ Since $\ddot s = \ddot x_M - \ddot x_m$ it follows that $$\ddot s = \frac FM - k\frac{M + m}{Mm}s = \frac FM - \omega^2 s.$$ Multiplying both sides by $\dot s$ and integrating with respect to time using $s(0) = \dot s(0) = 0$ then yields $$\dot s^2 + \omega^2 s^2 = \frac{2F}{M} s$$ from which the result follows.

 

[PeroK]

right after the instant when both blocks acquire the same velocity(i.e when there is maximum extension in the spring) won't both blocks from then on also have the same acceleration?

I don't see that. Why would that be?

 

[Hamiltonian]

I don't see that. Why would that be?

because both the blocks are constrained to have the same acceleration when the string has maximum extension and as the string stays that way from the point they attain the same velocity they will continue with the same acceleration?

 

[PeroK]

because both the blocks are constrained to have the same acceleration when the string has maximum extension and as the string stays that way from the point they attain the same velocity they will continue with the same acceleration?

In the CoM frame each block will execute harmonic motion (of some description). The zero acceleration will precede the zero velocity for block $m$ (in the CoM frame).

 

[Hamiltonian]

There is an alternative method which uses the laboratory frame.

If the initial length of the spring is $d$, then the extension is $s = x_M - x_m - d$ where $x_M$ is the position of mass $M$, etc.

By considering the forces on each mass in an inertial frame, you can obtain $$\begin{align*} M\ddot x_M &= F - ks \\ m\ddot x_m &= ks \end{align*}$$ Since $\ddot s = \ddot x_M - \ddot x_m$ it follows that $$\ddot s = \frac FM - k\frac{M + m}{Mm}s = \frac FM - \omega^2 s.$$ Multiplying both sides by $\dot s$ and integrating with respect to time using $s(0) = \dot s(0) = 0$ then yields $$\dot s^2 + \omega^2 s^2 = \frac{2F}{M} s$$ from which the result follows.

is there a special significance of $\omega$ that you are using?
also how exactly are you getting
$\dot s^2 + \omega^2 s^2 = \frac{2F}{M} s$. I am not extremely familiar with solving differential equations.

 

[haruspex]

is there a special significance of $\omega$ that you are using?
also how exactly are you getting
$\dot s^2 + \omega^2 s^2 = \frac{2F}{M} s$. I am not extremely familiar with solving differential equations.

@pasmith is merely writing $k\frac{M+m}{Mm}$ as $\omega^2$.

 

[Hamiltonian]

@pasmith is merely writing $k\frac{M+m}{Mm}$ as $\omega^2$.

yea I got that but why $\omega^2$ and not just x or y. But anyway I don't understand how @pasmith gets the second eqn

 

[haruspex]

yea I got that but why $\omega^2$ and not just x or y

Because, from experience, we will end up with the square root of that being an oscillation frequency.

 

[haruspex]

Then following the steps @pasmith outlined:

$$\ddot s = \frac FM - k\frac{M + m}{Mm}s = \frac FM - \omega^2 s.$$ Multiplying both sides by $\dot s$ and integrating with respect to time using $s(0) = \dot s(0) = 0$ then yields $$\dot s^2 + \omega^2 s^2 = \frac{2F}{M} s$$

$$\ddot s\dot s = \frac FM\dot s - \omega^2 \dot s s.$$
You can check that the integral of $\ddot s\dot s$ wrt t is $\frac 12(\dot s)^2$, etc.

 

[PeroK]

@Hamiltonian299792458

Alternatively, in the CoM frame:

1) There is a ficticious force on $m$ calculated from the acceleration of the CoM.

2) The displacement of $m$ from the CoM is a fixed proportion of the total extension of the spring. This results in a modified effective spring constant for $m$'s motion.

3) $m$ executes SHM as a hanging mass would under gravity associated with the fictitious force and the modified spring constant.

4) $M$ excutes SHM as a hanging mass under a different fictitious (gravity) force (includes real force $F$) and another modified spring constant.

 

[Hamiltonian]

I wanted to visualize the motion of the two blocks in the inertial frame.
Intuitively I felt that M would start moving first and as it does so it will drag m with it, and the distance between them will just keep on increasing till the time there is maximum extension in the spring.

So when you say m and M execute SHM, I think maybe I can visualize m doing so but M I don't see why it should execute SHM.

 

[PeroK]

So when you say m and M execute SHM, I think maybe I can visualize m doing so but M I don't see why it should execute SHM.

That's where mathematics can help you.

In any case, if (in the CoM frame) mass $m$ is oscillating with an angular frequency $w$, then mass $M$ must be oscillating in phase with the same angular frequency.

 

[PeroK]

@Hamiltonian299792458 here's my solution using CoM frame:

Spoiler

Let $x$ be the displacement (to the left) of mass $m$ and $X$ be the displacement (to the right) of mass $M$. Both relative to the CoM. The extension of the spring is $x + X$. We have:
$$mx = MX$$
In this frame we have an effective gravity acting to the left of magnitude:
$$g_e = a_{CoM} = \frac{F}{M+m}$$
We also have modified effective spring constants in this frame:
$$F_m = -k(x + X) = -k(\frac{M+m}{M})x = -k_e x \ \ \text{and} \ \ F_M = -k(x + X) = -k(\frac{M+m}{m})x = -K_e X$$
In this frame, mass $m$ is subject to effective gravity $g_e$ and effective spring constant $k_e$, hence executes SHM with angular frequency and amplitude:
$$w = \sqrt{\frac{k_e}{m}} = \sqrt{\frac{k(M+m)}{Mm}}, \ \ \text{and} \ \ A_m = \frac{mg_e}{k_e} = \frac{mMF}{k(M+m)^2}$$
And, using the relationship between $x$ and $X$, mass $M$ executes SHM with the same frequency and amplitude: $$A_M = \frac{m^2F}{k(M+m)^2}$$
The maximum extension is given by: $$2(A_m + A_M) = \frac{2mF}{k(M+m)}$$
We can double-check this by looking at the forces on $M$ in this frame. We have an angular frequency of:
$$w = \sqrt{\frac{K_e}{M}} = \sqrt{\frac{k(M+m)}{Mm}}$$
Which agrees with the above. And, we have an effective gravity (to the right) on $M$ of:
$$G_e = \frac F M - g_e$$
Hence an amplitude of:
$$A_M = \frac{MG_e}{K_e} = \frac{m^2F}{k(M+m)^2}$$
Which, again, agrees with the previous calculation.

 

[pasmith]

I wanted to visualize the motion of the two blocks in the inertial frame.
Intuitively I felt that M would start moving first and as it does so it will drag m with it, and the distance between them will just keep on increasing till the time there is maximum extension in the spring.

So when you say m and M execute SHM, I think maybe I can visualize m doing so but M I don't see why it should execute SHM.

In the inertial frame, the net force on the system is $F$. So on average, everything is being accelerated to the right. If you cancel that out by using the center of mass frame, then the masses do oscillate.