Question

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. 2 Hz
• B. 1.5 Hz
• C. 0.5 Hz
• D. 1 Hz

Answer 1

The Correct Option is D

To determine the frequency of oscillation for a particle executing simple harmonic motion (SHM) with the given conditions, we begin by using the formula for maximum speed in SHM.

The maximum speed (v_{\text{max}}) in SHM is given by the formula:

v_{\text{max}} = A \omega

where:

• A is the amplitude of the motion.
• \omega is the angular frequency.

We need to find the angular frequency \omega, which is related to the frequency f by:

\omega = 2\pi f

Substitute \omega = 2\pi f into the formula for v_{\text{max}}:

v_{\text{max}} = A \cdot (2\pi f)

Given: A = 5 \text{ cm} = 0.05 \text{ m} and v_{\text{max}} = 31.4 \text{ cm/s} = 0.314 \text{ m/s}.

Substitute these values into the equation:

0.314 = 0.05 \cdot 2 \pi f

Rearrange to solve for f:

f = \frac{0.314}{0.05 \cdot 2 \pi}

Calculate the frequency:

f = \frac{0.314}{0.1 \pi} = \frac{0.314}{0.314} = 1 \text{ Hz}

The frequency of the oscillation is 1 \text{ Hz}.

Conclusion: The frequency of the oscillation is 1 Hz. Therefore, the correct option is:

• 1 Hz