If a string fixed at both ends having fundamental frequency of `240 Hz` is vibrated with the help of a tuning fork having frequency `280 Hz `, then the

Options

A. string will vibrate with a frequency of `240 Hz`
B. string will be in resonance with the tuning fork
C. string will vibrate with the frequency of tuning fork , but resonance condition will not be achieved
D. string will vibrate with a frequency of `260 Hz`.

Explanation

To solve the problem, we need to analyze the situation involving the string and the tuning fork.

### Step-by-Step Solution:

1. **Identify the Fundamental Frequency of the String:**
The fundamental frequency of the string is given as \( f_s = 240 \, \text{Hz} \).

2. **Identify the Frequency of the Tuning Fork:**
The frequency of the tuning fork is given as \( f_t = 280 \, \text{Hz} \).

3. **Check for Resonance Condition:**
For resonance to occur, the frequency of the tuning fork must be a multiple of the fundamental frequency of the string. This means we need to check if \( f_t \) can be expressed as \( n \cdot f_s \), where \( n \) is an integer.

\[ \frac{f_t}{f_s} = \frac{280}{240} = \frac{7}{6} \approx 1.1667 \]

Since \( \frac{7}{6} \) is not an integer, the tuning fork frequency is not a multiple of the string's fundamental frequency.

4. **Conclusion about the Vibrations:**
Since the tuning fork frequency is not a multiple of the string's fundamental frequency, the string will vibrate at the frequency of the tuning fork, but it will not be in resonance.

5. **Select the Correct Option:**
Based on the analysis, the correct option is that the string will vibrate with the frequency of the tuning fork, but resonance condition will not be achieved.

### Final Answer:
The string will vibrate with the frequency of the tuning fork (280 Hz), but resonance condition will not be achieved.

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