- #1
decentfellow
- 130
- 1
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!