Question

A block is resting on a piston which executes simple harmonic motion with a period 2.0 s the maximum velocity of the piston, at amplitude just sufficient for the block to separate from the piston is?
A. $1.57$ $m{s^{ - 1}}$
B. $3.12$ $m{s^{ - 1}}$
C. $2.0$ $m{s^{ - 1}}$
D. $6.42$ $m{s^{ - 1}}$

Answer 1

As the block is doing simple harmonic motion it will attain the maximum velocity when the block is at minimum displacement from mean position, we are going to find velocity, by the fact that block will attain maximum acceleration at a position of maximum displacement.
Formula used:
$a = - {\omega ^2}x$
$v = \omega \sqrt {{A^2} - {x^2}}$
$\omega = \dfrac{{2\pi }}{T}$
Where $a$ is the acceleration of the block.
$\omega$ is the angular frequency of SHM.
$A$ is the amplitude of SHM
$x$ is the displacement of the block.
$T$ is the time period of SHM.

Complete answer:
When a piston performs SHM, the block resting on it also performs SHM but as the block and piston are not connected by anything the block can lose contact with the piston. The block will lose contact with the piston when the normal force between block and piston goes zero, then the only force acting on the block will be its own weight, therefore maximum possible acceleration of the block can be $g$.
Now we know that maximum acceleration can be $g$, so if the block achieves this much acceleration normal will get zero and contact between piston and block will lose. Also the maximum acceleration in any SHM occurs at an extreme position i.e. when $x = A$ , where $A$ is amplitude of SHM. Then,
\begin{array}{*{20}{c}} {a = {\omega ^2}x} \\\ {g = {\omega ^2}A} \\\ {A = \dfrac{g}{{{\omega ^2}}}} \end{array}
Now maximum velocity will happen at mean position i.e. at $x = 0$
So, \begin{array}{*{20}{l}} {v = \omega \sqrt {{A^2} - {0^2}} } \\\ {v = \omega A} \end{array}
But, above we calculated $A = \dfrac{g}{{{\omega ^2}}}$ , on substituting in equation for velocity we get,
$v = \dfrac{g}{\omega }$
Also
\begin{array}{*{20}{l}} {\omega = \dfrac{{2\pi }}{T}} \\\ {\omega = \dfrac{{2 \times 3.14}}{2}} \\\ {\omega = 3.14\;{\mkern 1mu} {\text{rad}}{\mkern 1mu} {{\text{s}}^{{\text{ - 1}}}}} \end{array}
So maximum velocity of block can be
\begin{array}{*{20}{l}} {v = \dfrac{{9.81}}{{3.14}}} \\\ {v = 3.12{\mkern 1mu} \;m{s^{ - 1}}} \end{array}
So the maximum velocity for which the block and piston cannot separate is$3.12{\mkern 1mu} \;m{s^{ - 1}}$.

Thus, option (B) is correct.

Note:
In a simple harmonic motion the body performing SHM has maximum velocity at mean position and minimum velocity at extreme position, in the same way any body executing simple harmonic motion achieves maximum acceleration at extreme position and minimum acceleration at mean position, because acceleration is directly proportional to displacement of body from mean position.