A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is

A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is:

(a) 4 Hz

(b) 3 Hz

(c) 2 Hz

(d) 1 Hz

Detailed Answer:

The particle is undergoing Simple Harmonic Motion (SHM). The relationship between maximum speed \(v_{\text{max}}\), angular frequency \(\omega\), and amplitude \(A\) in SHM is given by: \[ v_{\text{max}} = \omega A. \] Here, the amplitude \(A\) is 5 cm, and the maximum speed \(v_{\text{max}}\) is 31.4 cm/s. Substituting these values: \[ 31.4 = \omega \cdot 5. \] Solving for \(\omega\): \[ \omega = \frac{31.4}{5} = 6.28 \text{ rad/s}. \] The angular frequency \(\omega\) is related to the frequency \(f\) by: \[ \omega = 2\pi f \quad \Rightarrow \quad f = \frac{\omega}{2\pi}. \] Substituting \(\omega = 6.28\): \[ f = \frac{6.28}{2\pi} = 1 \text{ Hz}. \] Therefore, the correct option is: \[ \boxed{1 \text{ Hz} \ (\text{Option (d)})}. \] This calculation shows the frequency of oscillation based on the given amplitude and maximum speed.

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