A body of mass 8 kg performs S.H.M. of amplitude 60 cm. The restoring force is 120 N, when the displacement is 60 cm. The time period is

Options

A. `0.628 s`
B. `1.256 s`
C. `1.884 s`
D. ` 2.5 12s`

Answer

To find the time period of a body performing Simple Harmonic Motion (S.H.M.), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the body, \( m = 8 \, \text{kg} \)
- Amplitude, \( A = 60 \, \text{cm} = 0.6 \, \text{m} \) (convert to meters)
- Restoring force, \( F = 120 \, \text{N} \)
- Displacement, \( x = 60 \, \text{cm} = 0.6 \, \text{m} \)

2. **Use the Restoring Force Formula:**
The restoring force in S.H.M. is given by:
\[ F = -kx \]
where \( k \) is the spring constant. We can ignore the negative sign since we are interested in the magnitude:
\[ k = \frac{F}{x} \]
Substituting the known values:
\[ k = \frac{120 \, \text{N}}{0.6 \, \text{m}} = 200 \, \text{N/m} \]

3. **Relate Spring Constant to Angular Frequency:**
The relationship between spring constant \( k \), mass \( m \), and angular frequency \( \omega \) is given by:
\[ k = m\omega^2 \]
Rearranging for \( \omega^2 \):
\[ \omega^2 = \frac{k}{m} \]
Substituting the values:
\[ \omega^2 = \frac{200 \, \text{N/m}}{8 \, \text{kg}} = 25 \, \text{rad}^2/\text{s}^2 \]

4. **Calculate Angular Frequency \( \omega \):**
Taking the square root to find \( \omega \):
\[ \omega = \sqrt{25} = 5 \, \text{rad/s} \]

5. **Find the Time Period \( T \):**
The time period \( T \) is related to angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{5} \approx 1.256 \, \text{s} \]

6. **Final Answer:**
The time period \( T \) is approximately:
\[ T \approx 1.26 \, \text{s} \]