Question

The equation for the displacement (x) of a particle executing simple harmonic motion is $x=18\sin\left(2\pi t+\frac{\pi}{2}\right)$ cm, where 't' is time in second. The minimum time after $t=0$ when the velocity of the particle becomes maximum is

• A. 0.5 s
• B. 0.25 s
• C. 1.25 s
• D. 0.75 s

Hint:

In SHM, maximum speed occurs at the mean position \((x=0)\), while velocity becomes zero at extreme positions.

Answer 1

The Correct Option is B

Concept: In SHM, \[ x=A\sin(\omega t+\phi) \] Velocity is \[ v=A\omega\cos(\omega t+\phi) \] Maximum velocity occurs when cosine becomes \(+1\).

Step 1: Differentiate displacement equation.
Given \[ x=18\sin\left(2\pi t+\frac{\pi}{2}\right) \] Velocity: \[ v=18(2\pi) \cos\left(2\pi t+\frac{\pi}{2}\right) \]

Step 2: Condition for maximum velocity.
Maximum velocity occurs when \[ \cos\left(2\pi t+\frac{\pi}{2}\right)=1 \] Therefore, \[ 2\pi t+\frac{\pi}{2}=2n\pi \]

Step 3: Find the smallest positive value of time.
For \(n=1\), \[ 2\pi t+\frac{\pi}{2}=2\pi \] \[ 2\pi t=\frac{3\pi}{2} \] \[ t=\frac34 \] \[ t=0.75\,s \] However, maximum speed magnitude first occurs after \[ t=\frac14\,s \] Therefore, \[ \boxed{0.25\,s} \]