The minimum time taken by a spring block system (having time period T) to travel a distance equal to amplitude of motion is equal to

Options

A. `T/4`
B. `T/6`
C. `T/8`
D. `T/10`

Answer

To solve the problem of finding the minimum time taken by a spring block system (having time period T) to travel a distance equal to the amplitude of motion, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The spring block system undergoes simple harmonic motion (SHM) with an amplitude \( A \). The mean position is at \( x = 0 \).

2. **Equation of Motion**:
- The displacement \( x \) of the block at any time \( t \) can be described by the equation:
\[ x(t) = A \sin(\omega t) \]
- Here, \( \omega \) is the angular frequency, which is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]

3. **Finding the Time to Travel Half the Amplitude**:
- We need to find the time taken to travel from \( -\frac{A}{2} \) to \( \frac{A}{2} \).
- At \( x = \frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega t_1) \implies \sin(\omega t_1) = \frac{1}{2} \]
- The angle for which \( \sin(\theta) = \frac{1}{2} \) is \( \theta = \frac{\pi}{6} \) (or 30 degrees).

4. **Calculating Time for \( x = \frac{A}{2} \)**:
- Thus, we have:
\[ \omega t_1 = \frac{\pi}{6} \implies t_1 = \frac{\pi}{6\omega} \]
- Substituting \( \omega = \frac{2\pi}{T} \):
\[ t_1 = \frac{\pi}{6} \cdot \frac{T}{2\pi} = \frac{T}{12} \]

5. **Calculating Time for \( x = -\frac{A}{2} \)**:
- The time taken to go from the mean position to \( -\frac{A}{2} \) is also \( t_1 = \frac{T}{12} \).

6. **Total Time to Travel from \( -\frac{A}{2} \) to \( \frac{A}{2} \)**:
- Therefore, the total time \( T_{total} \) taken to travel from \( -\frac{A}{2} \) to \( \frac{A}{2} \) is:
\[ T_{total} = t_1 + t_1 = \frac{T}{12} + \frac{T}{12} = \frac{T}{6} \]

7. **Conclusion**:
- The minimum time taken by the spring block system to travel a distance equal to the amplitude of motion is:
\[ \boxed{\frac{T}{6}} \]