Question

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

• A. 2.047 s
• B. 1.047 s
• C. 3.047 s
• D. 0.047 s

Answer 1

Answer: (B)

1.047 s

Answer 2

Length of the line = Distance between extreme positions of oscillation = 4 cm So, Amplitude $a = 2cm.$ also

$v_{\max} = 12cm/s.$

$\because{v\frac{2\pi}{T}}_{\max}$

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