A mass `M` is suspended from a massless spring. An additional mass `m` stretches the spring further by a distance `x`. The combined mass will oscillate with a period

Options

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

Explanation

To find the time period of the oscillation of the combined mass suspended from a spring, we can follow these steps:

### Step 1: Understand the forces acting on the system
When the additional mass \( m \) is added to the mass \( M \), the total force acting on the spring is due to the weight of the combined mass, which is \( (M + m)g \), where \( g \) is the acceleration due to gravity.

### Step 2: Determine the spring constant \( k \)
The spring stretches by a distance \( x \) due to the additional mass \( m \). At equilibrium, the force exerted by the spring (which is \( kx \)) must balance the weight of the additional mass \( mg \):
\[ kx = mg \]
From this, we can express the spring constant \( k \):
\[ k = \frac{mg}{x} \]

### Step 3: Write the formula for the time period of oscillation
The formula for the time period \( T \) of a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m_{\text{total}}}{k}} \]
where \( m_{\text{total}} \) is the total mass attached to the spring.

### Step 4: Substitute the total mass and spring constant into the time period formula
The total mass \( m_{\text{total}} \) is \( M + m \). Substituting \( k \) from Step 2 into the time period formula gives:
\[ T = 2\pi \sqrt{\frac{M + m}{\frac{mg}{x}}} \]

### Step 5: Simplify the expression
This simplifies to:
\[ T = 2\pi \sqrt{\frac{(M + m)x}{mg}} \]

### Final Answer
Thus, the time period \( T \) of the oscillation of the combined mass is:
\[ T = 2\pi \sqrt{\frac{(M + m)x}{mg}} \]

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