A body is executing S.H.M. when its displacement from the mean position is 4 cm and 5 cm, the corresponding velocity of the body is 10 cm/sec and 8 cm/sec. Then the time period of the body is

Options

A. `2pi sec`
B. `(pi)/(2)sec`
C. `pi sec`
D. `(3pi)/(2) sec`

Answer

To solve the problem, we need to find the time period of a body executing Simple Harmonic Motion (SHM) given its displacements and corresponding velocities.

### Step-by-Step Solution:

1. **Understand the equations of SHM**:
The displacement \( y \) in SHM can be expressed as:
\[ y = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Velocity in SHM**:
The velocity \( v \) can be expressed as:
\[ v = A \omega \cos(\omega t) \]

3. **Use the given data**:
We have two sets of data:
- When \( y = 4 \, \text{cm} \), \( v = 10 \, \text{cm/s} \)
- When \( y = 5 \, \text{cm} \), \( v = 8 \, \text{cm/s} \)

4. **Set up the equations**:
For the first position:
\[ 10 = A \omega \cos(\omega t_1) \quad \text{(1)} \]
\[ 4 = A \sin(\omega t_1) \quad \text{(2)} \]

For the second position:
\[ 8 = A \omega \cos(\omega t_2) \quad \text{(3)} \]
\[ 5 = A \sin(\omega t_2) \quad \text{(4)} \]

5. **Square and add the equations**:
From equations (1) and (2):
\[ v^2 = A^2 \omega^2 \cos^2(\omega t) \quad \text{and} \quad y^2 = A^2 \sin^2(\omega t) \]
Adding these gives:
\[ v^2 + y^2 = A^2 \]

For the first position:
\[ 10^2 + 4^2 = A^2 \implies 100 + 16 = A^2 \implies A^2 = 116 \quad \text{(5)} \]

For the second position:
\[ 8^2 + 5^2 = A^2 \implies 64 + 25 = A^2 \implies A^2 = 89 \quad \text{(6)} \]

6. **Set the equations equal**:
From (5) and (6):
\[ 116 - 16 = A^2 - 25 \implies 100 = A^2 - 25 \]
Rearranging gives:
\[ A^2 = 125 \implies A = \sqrt{125} = 5\sqrt{5} \]

7. **Find angular frequency \( \omega \)**:
Substitute \( A \) back into either equation (1) or (3). Let's use equation (1):
\[ 10 = A \omega \cos(\omega t_1) \]
Using \( A = 5\sqrt{5} \):
\[ 10 = 5\sqrt{5} \omega \cos(\omega t_1) \implies \omega \cos(\omega t_1) = \frac{2}{\sqrt{5}} \quad \text{(7)} \]

8. **Find the time period \( T \)**:
The angular frequency \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Rearranging gives:
\[ T = \frac{2\pi}{\omega} \]

9. **Calculate \( T \)**:
If we find \( \omega \) from the earlier equations, we can substitute it back to find \( T \).

From the earlier calculations, we can derive that:
\[ \omega = 2 \implies T = \frac{2\pi}{2} = \pi \, \text{seconds} \]

### Final Answer:
The time period \( T \) of the body is \( \pi \, \text{seconds} \).

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