Question

A particle executes SHM of period $12$s. After two seconds, it passes through the centre of oscillation, the velocity is found to be $3.142cm{s}^{-1}$. The amplitude of oscillation is:

• A. $6$cm
• B. $3$ cm
• C. $24$cm
• D. $12$cm

Answer 1

Answer: (D)

$12$cm

Let $y=Asin\omega t$

Then $v=\frac{dy}{dt}=\omega Acos\omega t=\frac{2\pi }{T}Acos\frac{2\pi }{T}t$

$\therefore 3.142=\frac{2\times 3.142}{12}\times Acos\frac{2\pi }{12}\times 2$

$\therefore A=12cm$