A particle executes SHM of period $π$ and amplitude 2cm. Find the acceleration of particle when it is 1 cm from mean position.

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Solution

The correct option is B

$x = A s i n (ω t + ϕ_{0}) i s t h e g e n e r a l e q^{n} f o r S H M h e r e A = 2 c m$

$T = π s, ω = ω = \frac{2 π}{T} = 2 s^{7}$

$x = 2 s i n (2 t + ϕ_{0})$ .....(1)

$v = \frac{d x}{d t} = 4 c o s (2 t + ϕ_{0})$ ....(2)

$v = \frac{d^{2} x}{d t^{2}} = 8 s i n (2 t + ϕ_{0})$ ....(3)

Given that $x = 1 c m, f r o m (1) s i n (2 t + ϕ_{0}) = \frac{1}{2}$

from (3) $a = - 8 \times \frac{1}{2} = - 4 m s^{- 1}$ .

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A particle executes SHM of period π and amplitude 2cm. Find the acceleration of particle when it is 1 cm from mean position.

The correct option is B x=A sin(ω t+ϕ0) is the general eqn for SHM here A=2cm T=π s, ω=ω=2πT=2s7 x=2 sin(2t+ϕ0).....(1) v=dxdt=4 cos(2t+ϕ0)....(2) v=d2xdt2=8 sin(2t+ϕ0)....(3) Given that x=1cm,from(1)sin(2t+ϕ0)=12 from (3) a=−8×12=−4ms−1.

A particle executes SHM of period $π$ and amplitude 2cm. Find the acceleration of particle when it is 1 cm from mean position.