Question

An object of mass 3 kg is executing simple harmonic motion with an amplitude of \( \frac{2}{\pi} \, \text{m} \). If the kinetic energy of the object when it crosses the mean position is 6 J, the time period of oscillation of the object is:

• A. 1 s
• B. 2 s
• C. 3 s
• D. 4 s

Hint:

To find the time period of an oscillating system, use the relation \( T = \frac{1}{f} \), where \( f \) is the frequency. The frequency can be derived from the kinetic energy at the mean position.

Answer 1

The Correct Option is B

For simple harmonic motion (SHM), the total mechanical energy is given by the sum of the kinetic and potential energies. When the object crosses the mean position, its kinetic energy is maximum and potential energy is zero. The kinetic energy at the mean position is: \[ K.E = \frac{1}{2} m \omega^2 A^2 \] where: - \( m = 3 \, \text{kg} \) (mass), - \( A = \frac{2}{\pi} \, \text{m} \) (amplitude), - \( \omega = 2\pi f \) (angular frequency), - \( f \) is the frequency. Given that the kinetic energy at the mean position is 6 J, we can set up the equation: \[ 6 = \frac{1}{2} \times 3 \times (2\pi f)^2 \times \left( \frac{2}{\pi} \right)^2 \] Simplifying: \[ 6 = \frac{1}{2} \times 3 \times 4 \pi^2 f^2 \times \frac{4}{\pi^2} \] \[ 6 = 24 f^2 \] \[ f^2 = \frac{1}{4} \quad \Rightarrow \quad f = \frac{1}{2} \, \text{Hz} \] The time period \( T \) is the reciprocal of the frequency: \[ T = \frac{1}{f} = \frac{1}{\frac{1}{2}} = 2 \, \text{s} \] Thus, the time period of the oscillation is \( 2 \, \text{s} \), and the correct answer is option (2).

Answer 2

Step 1: Write down the given data
Mass, \( m = 3 \, \text{kg} \)
Amplitude, \( A = \frac{2}{\pi} \, \text{m} \)
Kinetic energy at mean position, \( K = 6 \, \text{J} \)

Step 2: Use the fact that at mean position, kinetic energy is maximum and equals total energy
Total mechanical energy, \( E = K = 6 \, \text{J} \)
Total energy in simple harmonic motion is given by:
\[ E = \frac{1}{2} k A^2 \]

Step 3: Find the spring constant \( k \)
\[ 6 = \frac{1}{2} k \left(\frac{2}{\pi}\right)^2 = \frac{1}{2} k \frac{4}{\pi^2} = \frac{2k}{\pi^2} \]
Multiply both sides by \( \pi^2 \):
\[ 6 \pi^2 = 2k \implies k = \frac{6 \pi^2}{2} = 3 \pi^2 \]

Step 4: Find angular frequency \( \omega \)
\[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{3 \pi^2}{3}} = \sqrt{\pi^2} = \pi \]

Step 5: Find time period \( T \)
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2 \, \text{seconds} \]

Final answer: Time period of oscillation is 2 seconds.