A string fixed at both ends whose fundamental frequency is `240 Hz` is vibrated with the help of a tuning fork having frequency `480 Hz`, then

Options

A. The string will vibrate with a frequency of `240 Hz`
B. The string will vibrate in resonance with the tuning fork
C. The string will vibrate in resonance with a frequency of `480 Hz`, but is not a resonance with the tuning fork
D. The string is in resonance with the tuning fork and hence vibrate with a frequency of `240 Hz`

Explanation

To solve the problem, we need to analyze the situation of a string fixed at both ends and its interaction with a tuning fork.

### Step-by-Step Solution:

1. **Identify the Fundamental Frequency**: The problem states that the fundamental frequency of the string is \( f_1 = 240 \, \text{Hz} \).

2. **Determine the Harmonics of the String**: For a string fixed at both ends, the frequencies of the harmonics can be calculated using the formula:
\[ f_n = n \cdot f_1 \]
where \( n \) is the harmonic number (1 for fundamental, 2 for first overtone, etc.).

- For \( n = 1 \): \( f_1 = 240 \, \text{Hz} \) (fundamental frequency)
- For \( n = 2 \): \( f_2 = 2 \cdot 240 = 480 \, \text{Hz} \) (first overtone)
- For \( n = 3 \): \( f_3 = 3 \cdot 240 = 720 \, \text{Hz} \) (second overtone)

3. **Identify the Frequency of the Tuning Fork**: The tuning fork has a frequency of \( f_{fork} = 480 \, \text{Hz} \).

4. **Check for Resonance**: Resonance occurs when the frequency of the tuning fork matches one of the harmonic frequencies of the string. In this case, \( f_{fork} = 480 \, \text{Hz} \) matches the first overtone frequency \( f_2 \).

5. **Conclusion**: Since the frequency of the tuning fork matches one of the harmonic frequencies of the string, the string will vibrate in resonance with the tuning fork at \( 480 \, \text{Hz} \).

### Final Answer:
The string will vibrate in resonance with the tuning fork at a frequency of \( 480 \, \text{Hz} \).

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