A horizontal spring mass system is executing SHM with time period of 4s. At time t=0, it is at mean position. Find the minimum time after which its potential energy becomes three times of kinetic energy.

Options

A. 1s
B. `1/2`s
C. `1/3`s
D. `2/3`s

Answer

To solve the problem step by step, we need to find the minimum time after which the potential energy of a horizontal spring-mass system executing Simple Harmonic Motion (SHM) becomes three times the kinetic energy.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Time period \( T = 4 \, \text{s} \)
- At \( t = 0 \), the mass is at the mean position (maximum displacement from equilibrium is zero).

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

3. **Write the Displacement Equation**:
- The displacement \( x \) at any time \( t \) can be expressed as:
\[ x = A \sin(\omega t) \]
- Since we start at the mean position, we can use \( \sin(\omega t) \).

4. **Potential and Kinetic Energy Expressions**:
- The potential energy \( PE \) in a spring is given by:
\[ PE = \frac{1}{2} k x^2 \]
- The kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2} m v^2 \]
- The velocity \( v \) can be expressed as:
\[ v = \frac{dx}{dt} = \omega A \cos(\omega t) \]

5. **Set Up the Relationship**:
- We need to find when \( PE = 3 \times KE \):
\[ \frac{1}{2} k x^2 = 3 \left(\frac{1}{2} m v^2\right) \]
- This simplifies to:
\[ k x^2 = 3 m v^2 \]

6. **Substituting Velocity**:
- Substitute \( v = \omega A \cos(\omega t) \):
\[ k x^2 = 3 m (\omega A \cos(\omega t))^2 \]

7. **Using \( k = m \omega^2 \)**:
- Since \( k = m \omega^2 \), we can substitute this into the equation:
\[ m \omega^2 x^2 = 3 m (\omega A \cos(\omega t))^2 \]
- Cancel \( m \) from both sides:
\[ \omega^2 x^2 = 3 \omega^2 A^2 \cos^2(\omega t) \]
- This simplifies to:
\[ x^2 = 3 A^2 \cos^2(\omega t) \]

8. **Expressing Displacement in Terms of Sine**:
- Since \( x = A \sin(\omega t) \):
\[ A^2 \sin^2(\omega t) = 3 A^2 \cos^2(\omega t) \]
- Dividing both sides by \( A^2 \):
\[ \sin^2(\omega t) = 3 \cos^2(\omega t) \]
- Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ \sin^2(\omega t) = 3 (1 - \sin^2(\omega t)) \]
- Let \( y = \sin^2(\omega t) \):
\[ y = 3(1 - y) \implies y + 3y = 3 \implies 4y = 3 \implies y = \frac{3}{4} \]
- Thus, \( \sin^2(\omega t) = \frac{3}{4} \).

9. **Finding the Angle**:
- Therefore, \( \sin(\omega t) = \frac{\sqrt{3}}{2} \).
- This corresponds to \( \omega t = \frac{\pi}{3} \) (or \( 60^\circ \)).

10. **Calculating Time**:
- Now substituting back for \( t \):
\[ \omega t = \frac{\pi}{3} \implies t = \frac{\pi}{3\omega} \]
- Substituting \( \omega = \frac{\pi}{2} \):
\[ t = \frac{\pi}{3 \cdot \frac{\pi}{2}} = \frac{2}{3} \, \text{s} \]

### Final Answer:
The minimum time after which the potential energy becomes three times the kinetic energy is \( \frac{2}{3} \, \text{s} \).

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