The amplitude and the periodic time of a S.H.M. are 5 cm and 6 sec respectively. At a distance of 2.5 cm away from the mean position, the phase will be

Options

A. `5 pi//12`
B. `pi//4`
C. `pi//3`
D. `pi//6`

Answer

To find the phase of a simple harmonic motion (S.H.M.) when the displacement is given, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Amplitude (A) = 5 cm
- Periodic Time (T) = 6 sec
- Displacement from mean position (x) = 2.5 cm

2. **Use the General Equation of S.H.M.:**
The general equation for S.H.M. can be expressed as:
\[ x = A \sin(\phi) \]
where \( \phi \) is the phase angle.

3. **Substitute Known Values:**
Substitute the known values into the equation:
\[ 2.5 = 5 \sin(\phi) \]

4. **Solve for \( \sin(\phi) \):**
Rearranging the equation gives:
\[ \sin(\phi) = \frac{2.5}{5} = \frac{1}{2} \]

5. **Determine the Phase Angle \( \phi \):**
The value of \( \phi \) for which \( \sin(\phi) = \frac{1}{2} \) can be found using the inverse sine function:
\[ \phi = \sin^{-1}\left(\frac{1}{2}\right) \]
The angle whose sine is \( \frac{1}{2} \) is:
\[ \phi = 30^\circ \quad \text{or} \quad \phi = \frac{\pi}{6} \text{ radians} \]

6. **Conclusion:**
Therefore, the phase \( \phi \) at a distance of 2.5 cm away from the mean position is:
\[ \phi = \frac{\pi}{6} \text{ radians} \]