A weightless spring of length 60 cm and force constant `100 N m^(-1)` is kept straight and unstretched on a smooth horizontal table and its ends are rigidly fixed. A mass of 0.25 kg is attached at the middle of the spring and is slightly displaced along the length. The time period of the oscillation of the mass is

Options

A. `pi/20` s
B. `pi/10` s
C. `pi/5` s
D. `pi/sqrt200` s

Answer

To find the time period of oscillation of a mass attached to a weightless spring that is fixed at both ends, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the parameters:**
- Length of the spring (L) = 60 cm = 0.6 m
- Spring constant (k) = 100 N/m
- Mass (m) = 0.25 kg

2. **Understanding the system:**
- The spring is fixed at both ends, and a mass is attached at the middle. When the mass is displaced slightly, it will oscillate due to the restoring force provided by the spring.

3. **Determine the effective spring constant:**
- When the spring is fixed at both ends and a mass is attached at the center, the spring can be considered as two springs in series, each with a spring constant of k.
- The effective spring constant (k_eff) for two springs in series is given by:
\[ k_{\text{eff}} = \frac{k}{2} \]
- Therefore, for our case:
\[ k_{\text{eff}} = \frac{100 \, \text{N/m}}{2} = 50 \, \text{N/m} \]

4. **Calculate the angular frequency (ω):**
- The angular frequency (ω) for a mass-spring system is given by:
\[ \omega = \sqrt{\frac{k_{\text{eff}}}{m}} \]
- Substituting the values:
\[ \omega = \sqrt{\frac{50 \, \text{N/m}}{0.25 \, \text{kg}}} = \sqrt{200 \, \text{s}^{-2}} = 10 \, \text{s}^{-1} \]

5. **Calculate the time period (T):**
- The time period (T) is related to the angular frequency by:
\[ T = \frac{2\pi}{\omega} \]
- Substituting the value of ω:
\[ T = \frac{2\pi}{10} = \frac{\pi}{5} \, \text{s} \]

### Final Answer:
The time period of the oscillation of the mass is:
\[ T = \frac{\pi}{5} \, \text{s} \]