First slide

Simple harmonic motion

Question

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

Moderate

A

2.047 s
B

1.047 s
C

3.047 s
D

0.047 s

Solution

Length of the line = Distance between extreme positions of oscillation = 4 cm
So, Amplitude $a = 2 cm .$
also $v_{\max} = 12 cm / s .$
$∵ v_{\max} = ωa = \frac{2 π}{T} a$
$\Rightarrow T = \frac{2 πa}{v_{\max}}$ $= \frac{2 \times 3 .14 \times 2}{12}$ $= 1 .047 \sec$

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