Question

Frequency of a particle executing $SHM$ is $10Hz$. The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Maximum speed of the particle is ($g=10m/s^2$)

• A. $2$$\pi m/s$
• B. $\pi m/s$
• C. $\frac{1}{\pi }m/s$
• D. $\frac{1}{2\pi }m/s$

Answer 1

Answer: (D)

$\frac{1}{2\pi }m/s$

Mean position of the particle is $mg/k$, distance below the unstretched position of spring. Therefore, amplitude of oscillation is $A=\frac{mg}{k}$

$\omega =\sqrt{\frac{k}{m}}=2\pi f=20\pi (f=10Hz)$

$\frac{m}{k}=\frac{1}{400{\pi }^2}$

$v_{max}=A\omega =\frac{g}{400{\pi }^2}\times 20\pi =\frac{1}{2\pi }m/s$