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. 100 Hz
B. 200 Hz
C. 400 Hz
D. 800 Hz

Answer

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

### Step 1: Identify the parameters
- Length of the string (L) = 1 m
- Mass of the string (m) = \(5 \times 10^{-4}\) kg
- Tension in the string (T) = 20 N
- Distance from one end where the string is plucked = 25 cm = 0.25 m

### Step 2: Calculate the mass per unit length (μ)
The mass per unit length (μ) of the string can be calculated using the formula:
\[ \mu = \frac{m}{L} \]
Substituting the values:
\[ \mu = \frac{5 \times 10^{-4} \text{ kg}}{1 \text{ m}} = 5 \times 10^{-4} \text{ kg/m} \]

### Step 3: Determine the fundamental frequency (f)
The fundamental frequency of a vibrating string fixed at both ends is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Substituting the values we have:
\[ f = \frac{1}{2 \times 1} \sqrt{\frac{20 \text{ N}}{5 \times 10^{-4} \text{ kg/m}}} \]

### Step 4: Simplify the expression
Calculating the expression inside the square root:
\[ \frac{20}{5 \times 10^{-4}} = 20 \times \frac{1}{5 \times 10^{-4}} = 20 \times 20000 = 400000 \]
Now, taking the square root:
\[ \sqrt{400000} = 200 \text{ Hz} \]

### Step 5: Calculate the frequency
Now substituting back into the frequency formula:
\[ f = \frac{1}{2} \times 200 = 100 \text{ Hz} \]

### Conclusion
The frequency of the stretched string when plucked at a point situated at 25 cm from one end is **100 Hz**.