Question

The piston in the cylinder head of a locomotive has a stroke of 6m. If the piston is executing simple harmonic motion with an angular frequency of $200\,rad\,{{\min }^{-1}}$ , its maximum speed is-
(A). $3m{{s}^{-1}}$
(B). $10m{{s}^{-1}}$
(C). $15m{{s}^{-1}}$
(D). $20m{{s}^{-1}}$

Answer 1

As the cylinder piston follows simple harmonic motion, it will oscillate around a mean or equilibrium position. The maximum distance it travels away from its mean position is its amplitude. Using the displacement, find an expression for velocity, the amplitude of velocity is its maximum value.

Formula used:
$y=A\sin \omega t$
$v=A\omega \cos \omega t$

Complete step by step solution:
Simple harmonic motion is a type of periodic motion in which the displacement is directly proportional to the restoring force but acts in the opposite direction towards an equilibrium position. Displacement in a simple harmonic motion is given by-
$y=A\sin \omega t$ - (1)
Here, $y$ is the displacement of the object from the equilibrium position
$A$ is the amplitude
$\omega$ is the angular velocity
$t$ is time taken
On differentiating eq (1) with respect to time, we get,
$\dfrac{dy}{dt}=A\omega \cos \omega t$
$\dfrac{dy}{dt}$ is velocity. Therefore,
$v=A\omega \cos \omega t$
From the above equation the maximum value of velocity will be-
${{v}_{\max }}=A\omega$ - (2)
The stroke is the total distance travelled by the cylinder piston, therefore
$A=\dfrac{stroke}{2}$

& \omega =200\,rad\,{{\min }^{-1}} \\\ & \Rightarrow \omega =\dfrac{200}{60}rad\,{{s}^{-1}} \\\ & \omega =3.3rad\,{{s}^{-1}} \\\ \end{aligned}$$ Substituting the value of angular velocity and given values in eq (1), we get $$\begin{aligned} & {{v}_{\max }}=\dfrac{6}{2}\times 3.3 \\\ & {{v}_{\max }}=9.9m\,{{s}^{-1}}\approx 10m{{s}^{-1}} \\\ \end{aligned}$$ The maximum velocity of the cylinder piston is about $$10m{{s}^{-1}}$$ , therefore, the correct option is (B). **So, the correct answer is “Option B”.** **Note:** Angular velocity is also called angular frequency and is defined as the rate of change of argument of the sine function. Simple harmonic motion varies according to sinusoidal function. The most common example of SHM is the oscillation of mass in spring. The displacement of spring is $$F=-kx$$ which means $$F\propto -x$$ .