in an experiment it was found that string vibrates in n loops when a mass M is placed on the pan. What mass should be placed on the pan to make it vibrate in 2n loops with same frequency ? ( neglect the mass of pan )

Options

A. `2M`
B. `M/4`
C. `4M`
D. `M/2`

Answer

To solve the problem, we need to understand the relationship between the mass placed on the string, the frequency of vibration, and the number of loops (or harmonics) formed. Here's a step-by-step solution:

### Step 1: Understand the relationship between mass and tension
The tension \( T \) in the string is directly related to the mass \( M \) placed on the pan. The tension can be expressed as:
\[ T = M \cdot g \]
where \( g \) is the acceleration due to gravity.

### Step 2: Relate frequency, tension, and linear mass density
The frequency \( f \) of a vibrating string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) is the length of the string,
- \( T \) is the tension,
- \( \mu \) is the linear mass density of the string.

### Step 3: Determine the initial conditions
When the mass \( M \) is placed on the pan, the string vibrates in \( n \) loops. The wavelength \( \lambda \) of the wave in the string can be expressed as:
\[ L = n \cdot \frac{\lambda}{2} \quad \Rightarrow \quad \lambda = \frac{2L}{n} \]
Substituting this into the frequency equation gives:
\[ f = \frac{n}{2L} \sqrt{\frac{M \cdot g}{\mu}} \]

### Step 4: Set up the condition for \( 2n \) loops
Now, we want the string to vibrate in \( 2n \) loops while maintaining the same frequency \( f \). The new wavelength \( \lambda' \) for \( 2n \) loops is:
\[ L = 2n \cdot \frac{\lambda'}{2} \quad \Rightarrow \quad \lambda' = \frac{L}{n} \]
Substituting this into the frequency equation gives:
\[ f = \frac{2n}{2L} \sqrt{\frac{M' \cdot g}{\mu}} \]
where \( M' \) is the new mass we need to find.

### Step 5: Equate the frequencies
Since we want the frequencies to be equal, we can set the two frequency equations equal to each other:
\[ \frac{n}{2L} \sqrt{\frac{M \cdot g}{\mu}} = \frac{2n}{2L} \sqrt{\frac{M' \cdot g}{\mu}} \]

### Step 6: Simplify the equation
Canceling common terms and simplifying gives:
\[ \sqrt{M} = 2\sqrt{M'} \]

### Step 7: Solve for the new mass \( M' \)
Squaring both sides results in:
\[ M = 4M' \quad \Rightarrow \quad M' = \frac{M}{4} \]

### Conclusion
To make the string vibrate in \( 2n \) loops with the same frequency, the mass that should be placed on the pan is:
\[ M' = \frac{M}{4} \]