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

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 3.142 cm s-1. The amplitude of oscillations is (A) 6cm (B) 3cm (C) 24cm (D) 12cm

Tardigrade Admin · Accepted Answer

Let y = A sin ω t Then v= (dy/dt) = ω A cosω t = (2π/T) = A cos (2π/T)t (∵ω = (2π/T)) ∴ 3.142 =( 2× 3.142/12)× A cos (2π/12)× 2 ∴ A= 12 cm

Q. A particle executes $S H M$ of period $12 s$ . After two seconds, it passes through the centre of oscillation, the velocity is found to be $3.142 c m s^{- 1}$ . The amplitude of oscillations is

$6$ cm

$3$ cm

$24$ cm

$12$ cm

Let $y = A s inω t$
Then $v = \frac{d y}{d t} = ω A cos ω t$
$= \frac{2 π}{T} = A cos \frac{2 π}{T} t (∵ ω = \frac{2 π}{T})$
$∴ 3.142 = \frac{2 \times 3.142}{12} \times A cos \frac{2 π}{12} \times 2$
$∴ A = 12 c m$

Q. A particle executes SHM of period 12s. After two seconds, it passes through the centre of oscillation, the velocity is found to be 3.142cms−1. The amplitude of oscillations is

6cm

3cm

24cm

12cm