Question

In an engine, the piston undergoes vertical simple harmonic motion with an amplitude of 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston is close to:
A. $0.7Hz$
B. $1.9Hz$
C. $1.2Hz$
D. $0.1Hz$

Answer 1

Hint: In the case of simple harmonic motion there will be a mean position and two extreme positions. The particle executes the SHM with respect to the mean position and within the range of two extreme positions. Generally, simple harmonic motions are denoted with the sinusoidal or cosecant functions.
Formula used:
$x = A\sin (\omega t)$

Complete step-by-step solution:
In SHM acceleration of the particle executing the SHM is proportional to the displacement of the particle from the mean position. Generally, if an object starts from the mean position then we write displacement as a sinusoidal function and if an object starts from the extreme position then we write the displacement of a body as a cosine function. We can assume in any way if it is not specified in the question. Let us assume that the given particle is executing the SHM about the mean position. Then the function of the SHM displacement will be $x = A\sin (\varphi )$ where $\varphi $ is the phase of the particle. That phase will be generally given as $\varphi = \omega t$ and A is the amplitude and $\omega = 2\pi f$ is the angular velocity and $f$ is the frequency.
So the displacement equation will become
$x = A\sin (\omega t)$
And the acceleration will be
$a = - {\omega ^2}x$
At the extreme position of the piston let it be the topmost position the acceleration acting will be
$a = - {\omega ^2}A$
The same pseudo acceleration will be acting on the washer as it is resting on the piston. Now if that upward pseudo acceleration equals the downwards acceleration due to gravity then the washer may lose contact with the piston.
So we have
$\eqalign{
  & {\omega ^2}A = g \cr
  & \Rightarrow {\left( {2\pi f} \right)^2}A = g \cr
  & \Rightarrow f = \sqrt {\dfrac{g}{{A4{\pi ^2}}}} \cr
  & \Rightarrow f = \sqrt {\dfrac{{10}}{{\left( {\dfrac{7}{{100}}} \right)4{\pi ^2}}}} \cr
  & \Rightarrow f = 1.9Hz \cr} $
Hence option B will be the answer.

Note: At the topmost point the acceleration will be acting downwards due to S.H.M but if we look from the piston frame of reference then the same amount of pseudo acceleration will be acting upwards on the washer. At the instant when that acceleration equals the downward acceleration due to gravity the washer will lose contact with the piston.