Maximum compression in the spring

[decentfellow]

Homework Statement

Find the maximum compression in the spring, if the lower block is shifted to rightwards with acceleration of '$a$'. All the surfaces are smooth.

Homework Equations

$$\vec{F}=m\vec{a}$$
$$\vec{F}_{sp}=k\vec{x}$$

The Attempt at a Solution

FBD of the upper block:

From the FBD of the upper block, we get
$$ma-kx=mv\dfrac{dv}{dx} \\
\implies \int_{0}^{v}{v dv} = \int_{0}^{x}{\left(a-\dfrac{k}{m}x\right)dx} \\
\implies \dfrac{v^2}{2}=ax-\dfrac{k}{2m}x^2$$

Now maximum compression in the spring will be at the moment when the speed of the block becomes $0$ the second time i.e. $x\neq 0$
So, on putting $v=0$, we get

$$ax-\dfrac{k}{2m}x^2=0 \implies x=0, \dfrac{2ma}{k}$$
$\therefore\qquad x=\dfrac{2ma}{k}$ is the maximum compression in the spring.

4. My deal with the question
Now, as you can you see that I have already solved the question, but after re-examining my solution I noticed that initially the block was in a position of rest i.e. $v=0$ and $x=0$. On substituting both of these in the very first differential equation we get $ma=0$. Now how is that possible because if that was to be the case then the block would not move at all and there would be no compression, hence there would be no point of posing the question. Now, I know I am definitely going wrong somewhere but I can't find where, so your help would be very much appreciated. Thanks, in advance!

 

[drvrm]

On substituting both of these in the very first differential equation we get ma=0ma=0ma=0. Now how is that possible because if that was to be the case then the block would not move at all and there would be no compression,

at time t=0 what is the state of motion of lower block?

 

[decentfellow]

at time t=0 what is the state of motion of lower block?

Isn't it moving with an acceleration $a$, but with zero velocity.

 

[TSny]

If you throw a ball straight upward, what is the ball's acceleration at the instant it reaches maximum height?

Does your answer agree with what you would expect from writing the acceleration as $a = v \frac{dv}{dy}$ and noting that $v = 0$ at max height? If not, can you see how to resolve the apparent paradox?

Hint: What is $v$ as a function of $y$ for the ball? Use this function to find $\frac{dv}{dy}$ at maximum height.

 

[TSny]

Your solution appears to use the frame of reference of the accelerating lower block. In your FBD, $ma$ is then the fictitious force that arises from being in an accelerating frame. If so, your solution appears to me to be correct. But note that the $a$ in $ma$ is not the acceleration of the block attached to the spring, it's the acceleration of your frame of reference relative to an inertial frame.

 

[decentfellow]

If you throw a ball straight upward, what is the ball's acceleration at the instant it reaches maximum height?

Does your answer agree with what you would expect from writing the acceleration as $a = v \frac{dv}{dy}$ and noting that $v = 0$ at max height? If not, can you see how to resolve the apparent paradox?

Hint: What is $v$ as a function of $y$ for the ball? Use this function to find $\frac{dv}{dy}$ at maximum height.

What I got from your post is that its essentially useless to define acceleration as $v\dfrac{dv}{dy}$ at the moment when the velocity is zero and is also about to change direction because at that instant $\dfrac{dv}{dy}=\infty$ and $v=0$, so we get $a=\infty\cdot 0$, essentially acceleration is not defined at this moment. But, I am having trouble understanding how to solve this paradox, am I supposed to just not consider $a$ as $v\dfrac{dv}{dy}$ at this instant or consider $a$ as $\dfrac{v}{\dfrac{dy}{dv}}$, this gives us a $\dfrac{0}{0}$ from at least something which can be solvable by L'Hôpital.

So, in my case, if $ma-kx=mv\dfrac{dv}{dx}$, then when $x=0$ and $v=0$, we get $v\dfrac{dv}{dx}=\dfrac{a-kx}{2v}=a$. Okay, to me it seems like I have solved it... or not? And seems like all that talk of the L'Hôpital was for naught.

 

[TSny]

What I got from your post is that its essentially useless to define acceleration as $v\dfrac{dv}{dy}$ at the moment when the velocity is zero and is also about to change direction because at that instant $\dfrac{dv}{dy}=\infty$ and $v=0$, so we get $a=\infty\cdot 0$,

Yes. But I would say that acceleration is defined as $a = \frac{dv}{dt}$. Then, you can show that $a = v\frac{dv}{dy}$at any instant where $\frac{dv}{dy}$ is defined.

essentially acceleration is not defined at this moment.

$a$ is defined at this momement, but $a$ cannot be expressed as $v\frac{dv}{dy}$ at this moment.

So, in my case, if $ma-kx=mv\dfrac{dv}{dx}$, then when $x=0$ and $v=0$, we get $v\dfrac{dv}{dx}=\dfrac{a-kx}{2v}=a$.

But note that $\dfrac{a-kx}{2v}$ is not defined at the moment when $v = 0$

In your solution, what you ended up using is essentially $v dv = a dx$, where $a$ here is the acceleration of the block of mass $m$, not the acceleration of the lower block.

$v dv = a dx$ is valid even at instants of time where $\frac{dv}{dx}$ is undefined.

 

[decentfellow]

Yes. But I would say that acceleration is defined as $a = \frac{dv}{dt}$. Then, you can show that $a = v\frac{dv}{dy}$at any instant where $\frac{dv}{dy}$ is defined.

$a$ is defined at this momement, but $a$ cannot be expressed as $v\frac{dv}{dy}$ at this moment.But note that $\dfrac{a-kx}{2v}$ is not defined at the moment when $v = 0$

In your solution, what you ended up using is essentially $v dv = a dx$, where $a$ here is the acceleration of the block of mass $m$, not the acceleration of the lower block.

$v dv = a dx$ is valid even at instants of time where $\frac{dv}{dx}$ is undefined.

Oh no, I just saw it now that I had written the expression for $v\dfrac{dv}{dx}$ wrong, it should be $a-kx$, I apologise for that. So, the acceleration of the block of mass $m$ has an acceleration of $a_m=v\dfrac{dv}{dx}=v\left(\dfrac{a-kx}{v}\right)$, and when $v=0 \text{ and } x=0$ then we get the acceleration of the block of mass $m$ as $a$. I think that my question is resolved now thanks for all your effort.