A body of mass m is suspended from vertical spring and is set into simple harmonic oscillations of time period T. Next the spring is fixed at one end on a smooth horizontal table and same body is attached at the other end. The body is pulled slightly and then released to produce horizontal oscillations of the spring. The time period of horizontal oscillations is

Options

A. T
B. 2T
C. `T/2`
D. Infinite as the spring does not oscillate

Answer

To solve the problem, we need to analyze the two cases of oscillations described: one with a vertical spring and the other with a horizontal spring.

### Step-by-Step Solution:

**Case 1: Vertical Spring**

1. **Understanding the Setup**:
- A mass \( m \) is suspended from a vertical spring. The weight of the mass \( mg \) stretches the spring to a new equilibrium position.
- The spring constant is denoted as \( k \).

2. **Finding the Extension**:
- At equilibrium, the spring force equals the weight of the mass:
\[ kx_0 = mg \]
- Here, \( x_0 \) is the extension of the spring from its natural length due to the weight of the mass.

3. **Setting Up the Equation of Motion**:
- When the mass is pulled down an additional distance \( x \) and released, the net force acting on the mass can be expressed as:
\[ F_{\text{net}} = k(x + x_0) - mg \]
- Since \( kx_0 = mg \), we can simplify:
\[ F_{\text{net}} = kx \]

4. **Applying Newton's Second Law**:
- According to Newton's second law:
\[ F_{\text{net}} = ma \]
- Therefore, we have:
\[ kx = ma \]
- This gives us the acceleration:
\[ a = \frac{k}{m}x \]

5. **Finding the Angular Frequency**:
- The equation \( a = -\omega^2 x \) leads to:
\[ \omega^2 = \frac{k}{m} \]
- Thus, the angular frequency \( \omega \) is:
\[ \omega = \sqrt{\frac{k}{m}} \]

6. **Calculating the Time Period**:
- The time period \( T \) is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

**Case 2: Horizontal Spring**

1. **Understanding the Setup**:
- The same mass \( m \) is now attached to a spring fixed at one end on a smooth horizontal table.

2. **Setting Up the Equation of Motion**:
- When the mass is pulled slightly and released, the only force acting on the mass is the spring force \( kx \).
- The normal force balances the weight, so we can ignore it in the horizontal motion.

3. **Applying Newton's Second Law**:
- The net force is:
\[ F_{\text{net}} = -kx \]
- This leads to:
\[ ma = -kx \]

4. **Finding the Angular Frequency**:
- Similar to the vertical case:
\[ a = -\frac{k}{m}x \]
- Thus, we have:
\[ \omega^2 = \frac{k}{m} \]

5. **Calculating the Time Period**:
- The time period for horizontal oscillations is:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

### Conclusion:
The time period of the horizontal oscillations is the same as that of the vertical oscillations:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

### Final Answer:
The time period of horizontal oscillations is \( T \).

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