physics
posted by Pronab .
A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. 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.

A=10cm
w=2PIf=2PI/T= PI/3
f(t)=Acos(wt+Phi)
f(0)=10cm*cos(0+phi)=5 >>>phi=PI/3 radians
f(t)=10cm*cos(wt+PI/3)
f'=10cm*w*sin(wt+PI/3)=velocity
f"=10cm*w^2*cos(PI*t/3+PI/3)=acceleration 
Q 2) A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal ?

X=(10cm)sin(2¦Ð/6st+¦Ð/6),~11cm s