Question

The displacement of a particle executing SHM is given by \(y = 5\sin\left(4t + \frac{\pi}{3}\right)\). If \(T\) is the time period and the mass of the particle is \(2g\), the kinetic energy of the particle when \(t = \frac{T}{4}\) is given by

• A. \(0.4\,J\)
• B. \(0.5\,J\)
• C. \(3\,J\)
• D. \(0.3\,J\)

Hint:

In SHM, velocity is maximum when displacement is zero, and \(v = A\omega \cos(\omega t + \phi)\).

Answer 1

The Correct Option is D

Step 1: Identify SHM parameters.
Given:
\[ y = 5\sin\left(4t+\frac{\pi}{3}\right) \]
So amplitude:
\[ A = 5 \]
Angular frequency:
\[ \omega = 4 \]
Step 2: Find time period.
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \]
So:
\[ t = \frac{T}{4} = \frac{\pi}{8} \]
Step 3: Find velocity at this time.
Velocity in SHM:
\[ v = \frac{dy}{dt} = A\omega \cos(\omega t + \phi) \]
So:
\[ v = 5 \cdot 4 \cos\left(4\cdot \frac{\pi}{8} + \frac{\pi}{3}\right) \]
\[ v = 20 \cos\left(\frac{\pi}{2} + \frac{\pi}{3}\right) \]
\[ v = 20 \cos\left(\frac{5\pi}{6}\right) \]
\[ \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} \Rightarrow v = 20\left(-\frac{\sqrt{3}}{2}\right) = -10\sqrt{3} \]
Step 4: Compute kinetic energy.
Mass \(m = 2g = 2 \times 10^{-3}kg\) (taking \(g = 10\,m\,s^{-2}\) as gram conversion implies \(2g = 0.002kg\)).
Kinetic energy:
\[ KE = \frac{1}{2}mv^2 \]
\[ KE = \frac{1}{2}(0.002)(100 \cdot 3) \]
\[ KE = 0.001 \times 300 = 0.3J \]
Final Answer: \[ \boxed{0.3\,J} \]