A loaded vertical spring executes simple harmonic oscillations with period of 4 s. The difference between the kinetic energy and potential energy of this system oscillates with a period of

Options

A. 2s
B. 1s
C. 8s
D. 4s

Answer

To solve the problem, we need to determine the period of oscillation of the difference between the kinetic energy (KE) and potential energy (PE) of a loaded vertical spring executing simple harmonic motion (SHM) with a given period of 4 seconds.

### Step-by-Step Solution:

1. **Understand the System**:
The system is a loaded vertical spring executing SHM with a period \( T = 4 \, \text{s} \).

2. **Identify the Angular Frequency**:
The angular frequency \( \omega \) is related to the period by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the given period:
\[ \omega = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{rad/s} \]

3. **Write the Displacement Equation**:
The displacement \( x \) of the mass on the spring can be expressed as:
\[ x = A \sin(\omega t) \]
where \( A \) is the amplitude of the motion.

4. **Calculate the Kinetic Energy (KE)**:
The velocity \( v \) of the mass is given by:
\[ v = \frac{dx}{dt} = A \omega \cos(\omega t) \]
The kinetic energy is:
\[ KE = \frac{1}{2} mv^2 = \frac{1}{2} m (A \omega \cos(\omega t))^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \]

5. **Calculate the Potential Energy (PE)**:
The potential energy stored in the spring is:
\[ PE = \frac{1}{2} k x^2 \]
Since \( k = m \omega^2 \), we can write:
\[ PE = \frac{1}{2} m \omega^2 (A \sin(\omega t))^2 = \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t) \]

6. **Find the Difference Between PE and KE**:
The difference between potential energy and kinetic energy is:
\[ PE - KE = \frac{1}{2} m A^2 \omega^2 \sin^2(\omega t) - \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \]
Factoring out the common terms:
\[ PE - KE = \frac{1}{2} m A^2 \omega^2 (\sin^2(\omega t) - \cos^2(\omega t)) \]
Using the identity \( \sin^2(\theta) - \cos^2(\theta) = -\cos(2\theta) \):
\[ PE - KE = -\frac{1}{2} m A^2 \omega^2 \cos(2\omega t) \]

7. **Determine the Period of the Difference**:
The term \( \cos(2\omega t) \) indicates that the oscillation of the difference between potential and kinetic energy has a frequency of \( 2\omega \). Therefore, the period \( T' \) of this oscillation is:
\[ T' = \frac{2\pi}{2\omega} = \frac{2\pi}{2 \cdot \frac{\pi}{2}} = 2 \, \text{s} \]

### Final Answer:
The period of oscillation of the difference between the kinetic energy and potential energy is \( 2 \, \text{s} \).

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