Question

Calculate the velocity of a particle performing S.H.M. after 1 second, if its displacement is given by \( x = 5 \sin\left(\frac{\pi t}{3}\right) \) m.

Hint:

In simple harmonic motion, velocity is the derivative of displacement with respect to time. The maximum velocity occurs when the displacement is zero.

Answer 1

Step 1: Given displacement equation.
The displacement of the particle is given by: \[ x = 5 \sin\left(\frac{\pi t}{3}\right) \, \text{m} \] where \( t \) is time.
Step 2: Find velocity equation.
The velocity \( v \) is the derivative of displacement with respect to time: \[ v = \frac{dx}{dt} = 5 \cdot \frac{\pi}{3} \cdot \cos\left(\frac{\pi t}{3}\right) \]
Step 3: Calculate velocity at \( t = 1 \, \text{s} \).
Substitute \( t = 1 \) second into the velocity equation: \[ v = 5 \cdot \frac{\pi}{3} \cdot \cos\left(\frac{\pi}{3}\right) \] Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \), we get: \[ v = \frac{5\pi}{3} \, \text{m/s} \]
Step 4: Conclusion.
The velocity of the particle after 1 second is \( \frac{5\pi}{3} \, \text{m/s} \).