Question

A $30g$ particle moves in simple harmonic motion with a frequency of $3$ oscillations per second and an amplitude of $5cm$. Calculate maximum speed $\left(inm/s\right)$ and maximum acceleration $\left(inm/s^2\right)$ of the particle respectively.

• A. $17.8,0.94$
• B. $0.94,0.94$
• C. $0.94,17.8$
• D. $17.8,17.8$

Answer 1

Answer: (C)

$0.94,17.8$

Given that the frequency of oscillation in SHM is $f=3\text{Hz}$

The displacement amplitude is $A=5\text{cm}$

A general SHM is represented by sinusoidal motion $x=A\sin \left(\omega t\right)$

Here, $\omega =2\pi f=6\pi \text{rad/s}$

Thus, velocity is give by $v=\frac{dx}{dt}=A\omega \cos \left(\omega t\right)$

and, acceleration is given by $a=\frac{dv}{dt}=-A{\omega }^2\sin \left(\omega t\right)$

Thus, maximum velocity is $v_{\text{max}}=A\omega =30\pi \text{cm/s}\approx 0.942\text{m/s}$

Similarly, maximum acceleration is given by $a_{\text{max}}=A{\omega }^2=180{\pi }^2{\text{cm/s}}^2\approx 17.76{\text{m/s}}^2$