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

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

Answer 1

The correct option is B

1.047 s

Length of the line = Distance between extreme

position of oscillation = 4 cm

So, Amplitude a = 2 cm.

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

$\because v_{max}=\omega a=\frac{2\pi }{T}a$

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