A particle executes SHM with a time period of 12s. Find the time taken by the particle to go directly from its mean position to half of its amplitude.

Answer

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

### Step-by-Step Solution:

1. **Understand the Parameters**:
- The time period (T) of the SHM is given as 12 seconds.
- The amplitude (A) is not specified but is not necessary for the calculation since we are looking for a ratio.

2. **Identify the Displacement**:
- The mean position corresponds to x = 0.
- Half of the amplitude is given by \( x = \frac{A}{2} \).

3. **Use the SHM Displacement Equation**:
- The standard equation for displacement in SHM is:
\[ x = A \sin(\omega t) \]
- Here, \( \omega \) (angular frequency) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
- Substituting the given time period:
\[ \omega = \frac{2\pi}{12} = \frac{\pi}{6} \text{ rad/s} \]

4. **Set Up the Equation for Half Amplitude**:
- We need to find the time \( t \) when the displacement is \( x = \frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega t) \]
- Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega t) \]

5. **Solve for \( \omega t \)**:
- Taking the inverse sine:
\[ \omega t = \sin^{-1}\left(\frac{1}{2}\right) \]
- We know that \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \), hence:
\[ \omega t = \frac{\pi}{6} \]

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

7. **Solve for \( t \)**:
- Cancel \( \frac{\pi}{6} \) from both sides:
\[ t = 1 \text{ second} \]

### Final Answer:
The time taken by the particle to go directly from its mean position to half of its amplitude is **1 second**.

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