In the engine of a locomotive, a cylindrical piece known as a piston oscillates in SHM in a cylinder head (cylindrical chamber) with an angular frequency of 180 rev/min. Its stroke (twice the amplitude) is 0.76 m.What is its maximum speed?

• Angular frequency of oscillation,$\omega =180\mathrm{rev}/\min$.
• Stroke of the piston = 2Maximum displacement of piston, $2x_{\mathrm{m}}=0.76\mathrm{m}$.

A piston of an engine of a locomotive executes a Simple Harmonic Motion. Using the formula of the maximum velocity of a body in oscillation, we can get the speed.

Formula:

The maximum velocity of a body in oscillation,$v_{max}=\omega x_m$ (i)

Conversion of rev/min to rad/sec,

$x\frac{\mathrm{rev}}{\min }=x\frac{\mathrm{rev}}{\min }\times \frac{1\min }{60s}\times \frac{2\pi \mathrm{rad}}{\mathrm{rev}}$ (ii)

The first step will be to convert the angular frequency into the units of rad/s from the given rev/min.

Using the formula from equation (ii), we get the angular frequency as:

$$\begin{gathered} \omega =180\frac{\mathrm{rev}}{\min } \\ =180\frac{\mathrm{rev}}{\min }\times \frac{1\min }{60\mathrm{s}}\times \frac{2\mathrm{\tau \tau }\mathrm{rad}}{\mathrm{rev}} \\ =6\mathrm{\tau \tau } \\ =18.8\mathrm{rad}/\mathrm{s} \end{gathered}$$

Also, the maximum displacement of the piston is given as:

$$\begin{gathered} {\mathrm{x}}_{\mathrm{m}}=\frac{0.76}{2}\left(\mathrm{from}\mathrm{the}\mathrm{given}\mathrm{data}\right) \\ =0.38\mathrm{m} \end{gathered}$$

For SHM,themaximum speed of oscillations using equation (i) and the given values is given by:

$$\begin{gathered} v_{\max }=18.8\times \times 0.38 \\ =7.2\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the maximum speed is 7.2 m/s .