A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6s. At t=0 it is at position x=5 cm 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

`y_(1) + y_(2) = 4 xx 10^(-3)[sin2pi(t - (x)/(2)) + sin2pi(t + (x)/(2))]`
At `t = 0` and `x = 0 , y_(1) + y_(2) = 0`
At `x = 2.333, y_(1) + y_(2) = 4 xx 10^(-3)m`
Attinodes at `x = 1,2`
Nodes at `x = (1)/(2), (1)/(3)`

Explanation

`y_(1) = 4 xx 10^(-3) sin 2pi (t - (x)/(2))`
`y_(2) = 4 xx 10^(3) sin 2pi(t + (x)/(2))`
`y_(1) + y_(2) = 4 xx 10^(-3) [sin2pi(t - (x)/(2)) + sin2pi(t + (x)/(2))]`
`= 8 xx 10^(-3) cos pix t`
At `t = 0`, and `x = 0, y_(1) + y_(2) = 0`
At `x = 2.333 , y_(1) + y_(2) = 8 xx 10^(-3) cos 0.333pi`
`= 4 xx 10^(-3) m`
Antinodes at `x = (1)/(2), (3)/(2)`