The vertical extension in a light spring by a weight of 1 kg suspended from the wire is 9.8 cm . The period of oscillation

Options

A. `20 pi ` cm
B. ` 2 pi `sec
C. `2 pi // 10 sec`
D. `200 pi sec`

Answer

To solve the problem of finding the period of oscillation for a spring extended by a weight, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass (m) = 1 kg
- Gravitational acceleration (g) = 9.8 m/s²
- Extension of the spring (x) = 9.8 cm = 0.098 m (convert cm to m)

2. **Calculate the Spring Constant (k):**
- According to Hooke's Law, the force exerted by the spring is equal to the weight of the mass at equilibrium:
\[ F = kx \]
- The weight (force due to gravity) can be calculated as:
\[ F = mg = 1 \text{ kg} \times 9.8 \text{ m/s}^2 = 9.8 \text{ N} \]
- Setting the two forces equal gives us:
\[ mg = kx \implies 9.8 = k \times 0.098 \]
- Rearranging to find k:
\[ k = \frac{9.8}{0.098} = 100 \text{ N/m} \]

3. **Calculate the Period of Oscillation (T):**
- The formula for the period of oscillation of a mass-spring system is:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
- Substituting the values of m and k:
\[ T = 2\pi \sqrt{\frac{1 \text{ kg}}{100 \text{ N/m}}} \]
- Simplifying the expression:
\[ T = 2\pi \sqrt{0.01} = 2\pi \times 0.1 = 0.2\pi \text{ seconds} \]
- Approximating π as 3.14:
\[ T \approx 0.2 \times 3.14 = 0.628 \text{ seconds} \]

### Final Answer:
The period of oscillation is approximately **0.628 seconds**.

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