A particle executing a simple harmonic motion has a period of 6 s. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

Options

A. `1/4s`
B. `3/4s`
C. `1/2s`
D. `3/2s`

Answer

To solve the problem of finding the time taken by a particle executing simple harmonic motion (SHM) to move from the mean position to half the amplitude, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the parameters**:
- The time period (T) of the SHM is given as 6 seconds.
- The amplitude (A) is not given explicitly, but we will denote it as A.

2. **Write the equation of motion**:
- The displacement of a particle in SHM can be expressed as:
\[ x(t) = A \sin(\omega t) \]
- Here, \(\omega\) is the angular frequency, which is related to the time period by the formula:
\[ \omega = \frac{2\pi}{T} \]

3. **Calculate the angular frequency**:
- Substituting the given time period:
\[ \omega = \frac{2\pi}{6} = \frac{\pi}{3} \text{ rad/s} \]

4. **Set up the equation for half the amplitude**:
- We want to find the time taken to move from the mean position (x = 0) to half the amplitude (x = A/2):
\[ A \sin(\omega t) = \frac{A}{2} \]
- Dividing both sides by A (assuming A ≠ 0):
\[ \sin(\omega t) = \frac{1}{2} \]

5. **Solve for \(\omega t\)**:
- The sine function equals \(\frac{1}{2}\) at \(\frac{\pi}{6}\) (30 degrees):
\[ \omega t = \frac{\pi}{6} \]

6. **Substitute for \(\omega\)**:
- We know \(\omega = \frac{\pi}{3}\), so:
\[ \frac{\pi}{3} t = \frac{\pi}{6} \]

7. **Solve for t**:
- Rearranging gives:
\[ t = \frac{\frac{\pi}{6}}{\frac{\pi}{3}} = \frac{1}{2} \text{ seconds} \]

8. **Conclusion**:
- The time taken by the particle to move from the mean position to half the amplitude is:
\[ t = \frac{1}{2} \text{ seconds} \]

### Final Answer:
The time taken by the particle to move from the mean position to half the amplitude is **0.5 seconds** or **1/2 seconds**.

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