A stretched string of length ` 1 m` fixed at both ends , having a mass of `5 xx 10^(-4) kg` is under a tension of `20 N`. It is plucked at a point situated at `25 cm` from one end . The stretched string would vibrate with a frequency of

Options

A. `400 Hz`
B. `100 Hz`
C. `200 Hz`
D. `256 Hz`

Explanation

To find the frequency of the stretched string that is fixed at both ends and plucked at a point, we can follow these steps:

### Step 1: Determine the mass per unit length (μ) of the string
The mass per unit length (μ) is calculated using the formula:
\[ \mu = \frac{m}{L} \]
Where:
- \( m = 5 \times 10^{-4} \) kg (mass of the string)
- \( L = 1 \) m (length of the string)

Substituting the values:
\[ \mu = \frac{5 \times 10^{-4}}{1} = 5 \times 10^{-4} \text{ kg/m} \]

### Step 2: Identify the tension (T) in the string
The tension in the string is given as:
\[ T = 20 \text{ N} \]

### Step 3: Determine the frequency of the fundamental mode
For a string fixed at both ends, the fundamental frequency (first harmonic) is given by:
\[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Where:
- \( L = 1 \) m
- \( T = 20 \) N
- \( \mu = 5 \times 10^{-4} \) kg/m

Substituting the values:
\[ f_1 = \frac{1}{2 \times 1} \sqrt{\frac{20}{5 \times 10^{-4}}} \]
\[ f_1 = \frac{1}{2} \sqrt{\frac{20}{5 \times 10^{-4}}} = \frac{1}{2} \sqrt{4 \times 10^5} = \frac{1}{2} \times 200 = 100 \text{ Hz} \]

### Step 4: Determine the frequency when plucked at 25 cm
When the string is plucked at a point that is 25 cm from one end, it creates a standing wave with two loops (antinodes) in the length of the string. The frequency of the second harmonic (first overtone) is given by:
\[ f_2 = 2f_1 \]
Substituting the value of \( f_1 \):
\[ f_2 = 2 \times 100 = 200 \text{ Hz} \]

### Final Answer
The frequency of the stretched string when plucked at 25 cm from one end is:
\[ \boxed{200 \text{ Hz}} \]