A spring elongated by length 'L' when a mass 'M' is suspended to it. Now a tiny mass 'm' is attached and then released, its time period of oscillation is -

Options

A. `2pisqrt(((M+m)l)/(Mg))`
B. `2pisqrt((ml)/(Mg))`
C. `2pisqrt(L//g)`
D. `2pisqrt((Ml)/((m+M)g))`

Explanation

To solve the problem, we need to find the time period of oscillation when a tiny mass \( m \) is attached to a spring that has already been elongated by a mass \( M \).

### Step-by-Step Solution:

1. **Understand the Initial Condition**:
When a mass \( M \) is suspended from the spring, it causes an elongation of the spring by a length \( L \). This means that the spring is in equilibrium under the force of gravity acting on mass \( M \).

2. **Apply Hooke's Law**:
According to Hooke's Law, the restoring force exerted by the spring is proportional to the extension. Mathematically, this can be expressed as:
\[ F = k \cdot x \]
where \( F \) is the restoring force, \( k \) is the spring constant, and \( x \) is the extension (which is \( L \) in this case).

3. **Set Up the Equation for Equilibrium**:
At equilibrium, the force due to the weight of mass \( M \) is balanced by the restoring force of the spring:
\[ M \cdot g = k \cdot L \]
Rearranging this gives us the spring constant \( k \):
\[ k = \frac{M \cdot g}{L} \]

4. **Consider the Additional Mass**:
Now, when a tiny mass \( m \) is added to the system, the total mass that will oscillate is \( M + m \).

5. **Use the Formula for Time Period**:
The time period \( T \) of oscillation for a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m_{\text{total}}}{k}} \]
Substituting the total mass \( m_{\text{total}} = M + m \) and the expression for \( k \):
\[ T = 2\pi \sqrt{\frac{M + m}{\frac{M \cdot g}{L}}} \]

6. **Simplify the Expression**:
Simplifying the expression gives:
\[ T = 2\pi \sqrt{\frac{(M + m) \cdot L}{M \cdot g}} \]

### Final Answer:
The time period of oscillation is:
\[ T = 2\pi \sqrt{\frac{(M + m) \cdot L}{M \cdot g}} \]