A load of mass `M` is attached to the bottom of a spring of mass `'M //3'` and spring constant 'K'. If the system is set into oscillation, the time period of oscillation is

Options

A. `(2pi)/(3) sqrt((10M)/(K))`
B. `4pi sqrt((M)/(3K))`
C. `4pi sqrt((M)/(K))`
D. `2pi sqrt((M)/(3K))`

Explanation

To find the time period of oscillation for the given system, we need to consider both the mass of the load and the mass of the spring itself. Here’s a step-by-step solution:

### Step 1: Identify the masses involved
- The mass of the load attached to the spring is \( M \).
- The mass of the spring is \( \frac{M}{3} \).

### Step 2: Calculate the effective mass
When a mass is attached to a spring, the oscillation is influenced by both the mass of the load and the mass of the spring. Therefore, we need to find the total effective mass \( M_{eff} \) that contributes to the oscillation:
\[ M_{eff} = M + \frac{M}{3} \]

### Step 3: Simplify the effective mass
To simplify \( M_{eff} \):
\[ M_{eff} = M + \frac{M}{3} = \frac{3M}{3} + \frac{M}{3} = \frac{4M}{3} \]

### Step 4: Use the formula for the time period of oscillation
The formula for the time period \( T \) of a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{M_{eff}}{K}} \]
Substituting the effective mass into the formula:
\[ T = 2\pi \sqrt{\frac{\frac{4M}{3}}{K}} \]

### Step 5: Simplify the time period expression
Now, simplify the expression for \( T \):
\[ T = 2\pi \sqrt{\frac{4M}{3K}} = 2\pi \cdot \frac{2\sqrt{M}}{\sqrt{3K}} = \frac{4\pi \sqrt{M}}{\sqrt{3K}} \]

### Final Result
Thus, the time period of oscillation for the system is:
\[ T = \frac{4\pi \sqrt{M}}{\sqrt{3K}} \]

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