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 time period of a mass-spring system in two different orientations: vertical and horizontal.

### Step-by-Step Solution:

1. **Understanding the Vertical Spring System**:
- When a mass \( m \) is attached to a vertical spring, it stretches the spring due to the force of gravity.
- The weight of the mass \( m \) is given by \( mg \), where \( g \) is the acceleration due to gravity.
- At equilibrium, the spring force \( kx_0 \) (where \( k \) is the spring constant and \( x_0 \) is the initial extension) balances the weight of the mass:
\[ mg = kx_0 \]
- The time period \( T \) of the vertical oscillation is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

2. **Transition to Horizontal Spring System**:
- Now, when the spring is fixed at one end on a smooth horizontal table, the mass is attached to the other end.
- When the mass is pulled slightly and released, it will oscillate horizontally.
- In this case, the only forces acting on the mass in the horizontal direction are the spring force and the inertia of the mass.

3. **Analyzing Horizontal Oscillation**:
- The spring force acting on the mass when displaced by a distance \( x \) is given by \( F = -kx \).
- According to Newton's second law, the net force is equal to mass times acceleration:
\[ F = ma \]
- Therefore, we can write:
\[ -kx = m \frac{d^2x}{dt^2} \]
- Rearranging gives us:
\[ \frac{d^2x}{dt^2} + \frac{k}{m}x = 0 \]
- This is the standard form of simple harmonic motion.

4. **Finding the Time Period of Horizontal Oscillation**:
- The angular frequency \( \omega \) for the horizontal oscillation is:
\[ \omega = \sqrt{\frac{k}{m}} \]
- The time period \( T' \) of the horizontal oscillation is given by:
\[ T' = 2\pi \sqrt{\frac{m}{k}} \]

5. **Comparing the Time Periods**:
- From our calculations, we see that:
\[ T' = T \]
- Thus, the time period of the horizontal oscillations is the same as that of the vertical oscillations.

### Conclusion:
The time period of the horizontal oscillations is equal to the time period of the vertical oscillations, which is \( T \).

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

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