`A` mass `M` is suspended from a light spring. An additional mass `m` added to it displaces the spring further by distance `x` then its time period is

Options

A. `T=2pisqrt((mg)/(x(M+m)))`
B. `T=2pisqrt(((M+m)x)/(mg))`
C. `T=(pi)/(2)sqrt((m)/(x(M+m)))`
D. `T=2pisqrt((M+m)/(mgx))`

Answer

To solve the problem, we need to determine the time period of a mass-spring system when an additional mass is added to it. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
Initially, a mass \( m \) is suspended from a light spring. When an additional mass \( M \) is added, the total mass becomes \( M + m \). The spring stretches by a distance \( x \).

### Step 2: Write the Formula for Time Period
The time period \( T \) of a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{M_{\text{total}}}{k}} \]
where \( M_{\text{total}} \) is the total mass of the system and \( k \) is the spring constant.

### Step 3: Determine the Total Mass
The total mass when the additional mass \( M \) is added is:
\[ M_{\text{total}} = m + M \]

### Step 4: Find the Spring Constant \( k \)
To find the spring constant \( k \), we can use Hooke's Law, which states that the force exerted by a spring is proportional to its extension:
\[ F_{\text{spring}} = kx \]
At equilibrium, the weight of the total mass equals the spring force:
\[ (M + m)g = kx \]
From this, we can express the spring constant \( k \):
\[ k = \frac{(M + m)g}{x} \]

### Step 5: Substitute \( k \) into the Time Period Formula
Now, substitute the expression for \( k \) back into the time period formula:
\[ T = 2\pi \sqrt{\frac{M + m}{\frac{(M + m)g}{x}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{M + m}{(M + m)g/x}} = 2\pi \sqrt{\frac{x}{g}} \]

### Final Answer
The time period of the system when an additional mass \( M \) is added and the spring is displaced by distance \( x \) is:
\[ T = 2\pi \sqrt{\frac{x}{g}} \]