A particle executes SHM in a line 4 cm long. Its velocity w…

Physics Simple Harmonic Motion (Oscillations) Time Period and Frequency Single Correct MCQ

Published on: September 12, 2026

A particle executes SHM in a line 4 cm long. Its velocity when passing through the centre of line is 12 cm/s. The period will be

2.047 s

1.047 s

3.047 s

0.047 s

Text Solution

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The correct answer is:

Length of the line = Distance between extreme positions of oscillation = 4 cm

So, Amplitude $a = 2 \mathrm{cm}.$

also $\mathbf{v}_{\max} = 12 \mathrm{cm}/\mathrm{s}.$

$\therefore v = \frac{2\pi}{T_{\max}}$ $\Rightarrow T = \frac{2\pi a}{v_{\max}} = \frac{2 \times 3.14 \times 2}{12} = 1.047 \text{ sec}$

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