A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand.
(a) What is the amplitude of oscillation ?
(b) Find the frequency of oscillation?

Options

A. 1 cm, 4.5Hz
B. 2 cm, 3.51Hz
C. 3 cm, 2.51Hz
D. 4 cm, 1.5Hz

Explanation

To solve the problem step by step, we will break it down into two parts: finding the amplitude of oscillation and finding the frequency of oscillation.

### Step-by-Step Solution

**Given:**
- The lowest position attained by the mass during oscillation is 4 cm below the initial position.
- The mass is attached to a massless spring.

**Part (a): Finding the Amplitude of Oscillation**

1. **Understanding the Setup:**
- The mass is initially held at a point where the spring is neither stretched nor compressed.
- When released, the mass oscillates and reaches a maximum extension of 4 cm below the initial position.

2. **Identifying the Amplitude:**
- The amplitude of oscillation (A) is defined as the maximum displacement from the mean position.
- Since the mass reaches 4 cm below the initial position, the amplitude of oscillation is:
\[ A = \frac{x}{2} = \frac{4 \, \text{cm}}{2} = 2 \, \text{cm} \]
- Thus, the amplitude of oscillation is **2 cm**.

**Part (b): Finding the Frequency of Oscillation**

1. **Using the Formula for Frequency:**
- The frequency (f) of oscillation is related to the mass (m) and spring constant (k) by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
- Where T is the time period. The frequency is given by:
\[ f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]

2. **Relating the Spring Constant to the Given Information:**
- From energy conservation, we have:
\[ m g x = \frac{1}{2} k x^2 \]
- Where \( x = 4 \, \text{cm} = 0.04 \, \text{m} \).
- Rearranging gives:
\[ k = \frac{2mg}{x} \]

3. **Substituting for Frequency:**
- Substitute \( k \) into the frequency formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{2mg}{x m}} = \frac{1}{2\pi} \sqrt{\frac{2g}{x}} \]
- Using \( g \approx 10 \, \text{m/s}^2 \) and \( x = 0.04 \, \text{m} \):
\[ f = \frac{1}{2\pi} \sqrt{\frac{2 \times 10}{0.04}} = \frac{1}{2\pi} \sqrt{500} \approx \frac{1}{2\pi} \times 22.36 \approx \frac{22.36}{6.28} \approx 3.56 \, \text{Hz} \]

### Final Answers:
- (a) Amplitude of oscillation: **2 cm**
- (b) Frequency of oscillation: **3.56 Hz**