Question

The displacement of a particle executing SHM is given by \( X = 0.01 \sin 100\pi (t + 0.05) \). The time period of the particle is:

• A. \(0.1 \, {s}\)
• B. \(0.01 \, {s}\)
• C. \(0.2 \, {s}\)
• D. \(0.02 \, {s}\)

Hint:

For SHM, always remember \(T = \frac{2\pi}{\omega}\), where \(\omega\) is the angular frequency.

Answer 1

The Correct Option is D

Step 1: Identify the angular frequency \(\omega\). The angular frequency \(\omega\) from the equation \(X = 0.01 \sin 100\pi (t + 0.05)\) is \(100\pi \, {rad/s}\). 
Step 2: Calculate the time period \(T\). The time period \(T\) of simple harmonic motion is given by \(T = \frac{2\pi}{\omega}\). \[ T = \frac{2\pi}{100\pi} = 0.02 \, {s} \]

Answer 2

To solve the problem, we need to determine the time period of the particle executing Simple Harmonic Motion (SHM) given its displacement equation.

1. Understanding the Displacement Equation:
The displacement equation is given by: \[ X = 0.01 \sin (100 \pi (t + 0.05)) \] In SHM, the general displacement equation is: \[ X = A \sin (\omega t + \phi) \] where: - \( A \) is the amplitude, - \( \omega \) is the angular frequency, - \( t \) is time, - \( \phi \) is the phase constant. From the given equation, we can compare the angular frequency \( \omega \) with the term in front of \( t \). Here, \( \omega = 100 \pi \) radians per second.

2. Finding the Angular Frequency and Time Period:
The angular frequency \( \omega \) is related to the time period \( T \) by the formula: \[ \omega = \frac{2\pi}{T} \] Given \( \omega = 100 \pi \), we can substitute into the formula: \[ 100 \pi = \frac{2 \pi}{T} \] Solving for \( T \), we get: \[ T = \frac{2 \pi}{100 \pi} = 0.02 \, \text{seconds} \]

3. Identifying the Correct Answer:
The time period \( T \) of the particle is 0.02 seconds, which matches Option 4.

Final Answer:
The correct answer is Option D: 0.02 s.