Question

If the displacement of a particle executing simple harmonic motion is given by \( x = 0.5 \cos(125.6\,t) \), then the time period of oscillation of the particle is nearly (Here \(x\) is displacement in metre and \(t\) is time in second)

• A. \(1 \, \text{s}\)
• B. \(2 \, \text{s}\)
• C. \(0.09 \, \text{s}\)
• D. \(0.05 \, \text{s}\)

Hint:

In SHM equations of the form \( x = A \cos(\omega t) \), the angular frequency \(\omega\) helps directly determine the time period using \( T = \frac{2\pi}{\omega} \).

Answer 1

The Correct Option is D

Step 1: Compare with standard SHM equation: \[ x = A \cos(\omega t) \Rightarrow \omega = 125.6 \, \text{rad/s} \] Step 2: Use the relation between angular frequency and time period: \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{125.6} \] Step 3: Calculate: \[ T \approx \frac{6.28}{125.6} \approx 0.05 \, \text{seconds} \] % Final Answer \[ \boxed{0.05 \, \text{s}} \]