Question

If the time period of a particle executing SHM is 8 s, then the time period of the potential energy of this particle is

• A. 16 s
• B. 4 s
• C. 2 s
• D. 8 s
• E. 32 s

Hint:

Energy completes two full cycles for every one cycle of displacement in SHM.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question deals with the properties of Simple Harmonic Motion (SHM). We need to understand how the potential energy of an oscillating particle varies with time and what its period is in relation to the period of the motion itself.
Step 2: Key Formula or Approach:
Let the displacement of a particle in SHM be given by: \[ x(t) = A \sin(\omega t) \] where \( T = \frac{2\pi}{\omega} \) is the period of the SHM. The potential energy (PE) of the particle (like in a spring-mass system) is given by: \[ PE(t) = \frac{1}{2}kx^2 = \frac{1}{2}k[A \sin(\omega t)]^2 = \frac{1}{2}kA^2 \sin^2(\omega t) \] We need to find the period of the function \( \sin^2(\omega t) \). Using the trigonometric identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \): \[ PE(t) = \frac{1}{2}kA^2 \left( \frac{1 - \cos(2\omega t)}{2} \right) = \frac{1}{4}kA^2 - \frac{1}{4}kA^2 \cos(2\omega t) \] Step 3: Detailed Explanation:
The displacement \( x(t) \) of the particle oscillates with an angular frequency \( \omega \). Its time period is \( T_{SHM} = \frac{2\pi}{\omega} \). The potential energy, as shown in the formula above, \( PE(t) \), oscillates due to the term \( \cos(2\omega t) \). The angular frequency of the potential energy oscillation is \( \omega_{PE} = 2\omega \). The time period of the potential energy (\( T_{PE} \)) is related to its angular frequency: \[ T_{PE} = \frac{2\pi}{\omega_{PE}} = \frac{2\pi}{2\omega} = \frac{1}{2} \left( \frac{2\pi}{\omega} \right) \] Since \( T_{SHM} = \frac{2\pi}{\omega} \), we have: \[ T_{PE} = \frac{1}{2} T_{SHM} \] The period of the potential energy is half the period of the simple harmonic motion. We are given that the time period of the particle's SHM is \( T_{SHM} = 8 \text{ s} \). Therefore, the time period of the potential energy is: \[ T_{PE} = \frac{1}{2} \times 8 \text{ s} = 4 \text{ s} \] Step 4: Final Answer:
The time period of the potential energy is 4 s.