The displacement time graph of a particle executing S.H.M. is shown in figure. Which of the following statement is `//` are true ?

Options

A. The force is zero at `t=(3T)/(4)`
B. The acceleration is maximum at `t=(4T)/(4)`
C. The velocity is maximum at `t=(T)/(4)`
D. The P.E. is equal to K.E. of oscillation at `t=(T)/(2)`

Answer

A, B, C

Explanation

For the given SHM, the displacement is given by `y=acos omegat`
Velocity, `V=(dy)/(dt)=-aomega sin omegat=aomega sin (omegat+pi)`
Force `=` mass `xx` acceleration `=-m a omega^(2) cos omega t`
Force is zero, when `cos omega t =0 or omega t =(pi)/(2) or (3pi)/(2), i.e., (2pi)/(T)t=(pi)/(2) or (3pi)/(2)`
If `(2pit)/(T)t=(pi)/(2),` then `t=(T)/(4)`
If `(2pi)/(T)t=(3pi)/(2), ` then `t=(3T)/(4)s` (given)
Acceleration is maximum if `cos omegat=1 or omega t=0 or 2pi or (2pi)/(T)t=2pi or t=T=(4T)/(4)s`
Velocity is maximum if `sin(omegat+pi)=1 or omegat+pi=pi//2`
or `omegat=(pi)/(2)-pi=-pi//2 or (2pi)/(T)t=-(pi)/(2) or t=-(T)/(4)s`
`PE=(1)/(2)momega^(2)y^(2)=(1)/(2)momega^(2)a^(2)cos^(2)omegat`
`KE=(1)/(2)momega^(2)a^(2)sin^(2)omegat`
If `PE=KE ` then `cos^(2)omegat=sin^(2)omegat=sin^(2)omegat or cosomegat=sinomegat or tanomegat=1`
or `omegat=(pi)/(4) or (2pi)/(T)t=(pi)/(4) or t=(T)/(8)s`