The displacement -time graph of a particle executing `SHM` is shown in figure. Which of the following statements 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 `PE` is equal to `KE` osc oscillation at `t = (T)/(2)`

Answer

A, B, C

Explanation

Displacement of the particle at any tome `t`,
`y = A_(0) cos omega t = A_(0) cos.(2pi)/(T)t`
and acceleration `A =- omega^(2)y = omega^(2) A_(0)cos .(2pi)/(T)t`.
In option (a) Firce `F = m omega^(2) y`.
At `t = (3T)/(4), y = A_(0) cos .((2pi)/(T)xx(3T)/(4))`
`=A_(0) cos .((3pi)/(2)) = 0`
Hence `F =0`, which makes option correct. In option (b) acceleration
`A =- omega^(2)y = omega^(2)A_(0) "cos"(2pi)/(T)t`.
`t = (4T)/(4),y = A_(0)cos .((2pi)/(T)xxT)`
`= A_(0) cos 2pi =+ A_(0)`
`:. a = omega^(2) A_(0)`, which is maximum value of `a`. Hence option (b) is correct.
In option (c ) The velocity `v = omega sqrt(A_(0)^(2)-y^(2))`
At `t = T//4`
`y = A_(0)cos ((2pi)/(T)xx(T)/(4)) = A_(0)cos((pi)/(2)) rArr y = 0`
Hence, `v = omegaA_(0)`, which is maximum value of velocity.
Hence option (c ) is correct.
In option (d) At `t = T//2`,
`y = A_(0) cos ((2pi)/(T)xx(T)/(2)) =- A_(0)`
Displacement is maximum, i.e corresponds to exterme position, it means `PE` is maximum and `KE` is zero.