Question

a particle executes shm with an amplitude of 10cm and time period 6s.At t=0 it is at position x=5cm going towards positive x-direction .Write the equation for the displacement x at time t.Find the magnitude of the acceleration of the particle at t=4s.

Answer 1

amplitude , A = 10cm
time period, T = 6 sec
we know, angular frequency , $\omega$ = 2π/T = 2π/6 = π/3 rad/sec.

equation of SHM, $x=Asin(\omega t+\Phi)$
at t = 0, x = 5

so, 5 = 10sin(π/3 × 0 + $\Phi$)
1/2 = sin$\Phi$
$\Phi$ = π/6

hence, equation of SHM, x = 10sin(π/3 t + π/6)

we know, acceleration , $a=-\omega^2x$
at t = 4sec ,
x = 10sin(π/3 × 4 + π/6)
= 10sin(4π/3 + π/6)
= 10sin(9π/6)
= 10sin(3π/2)
= -10cm = -0.1 m

so, acceleration , a = -(π/3)² × (-0.1) = π²/9 m/s²