Frequency of a particle executing SHM is 10 Hz. 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=10 m/s²)

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$π m / s$

$1 / 2 π m / s$

$1 / π m / s$

$2 π m / s$

answer is D.

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Detailed Solution

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

$\begin{array}{l} ω = \sqrt{\frac{k}{m}} = 2 π f = 20 π (f = 10 Hz) \\ \frac{m}{k} = \frac{1}{400 π^{2}} \\ v_{max} = A ω = \frac{g}{400 π^{2}} \times 20 π = \frac{1}{2 π} m / s \end{array}$

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