Question

The period of a simple pendulum is 2 second and the amplitude of vibration is 5 cm .Write down the equation of its motion What will be the displacement of the velocity of the particle after 1.5 second of the start of the motion

Answer 1

ans =33.33333...........

Answer 2

Given that time period of SHM = 2 sec = T

Amplitude = a = 5 cm

Angular velocity = w = 2π/T = 2π/2 = π

Displacement of particle in Simple Harmonic Motion is given by equation

X = a sin(wt + alpha)

Since the particle starts from mean position, alpha = 0

Hence equation of displacement is x = 5sin(πt)

Displacement at time = 1.5 sec is  

X = 5sin(π1.5) = 5sin(3π/2) = -5.

Hence the particle is 5 cm in the opposite direction.  

The scalar part of velocity is given by

V = w√(a² – x²) = π√(5*5 – (-5*-5)) = 0 cm/sec

Since the displacement is maximum, that means particle is having zero velocity to change the direction of SHM.