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Two instruments having stretched strings...

Two instruments having stretched strings are being played in unison . When the tension in one of the instruments is increases by ` 1% , 3` beats are produced in ` 2 s`. The initial frequency of vibration of each wire is

A

`600 Hz`

B

`300 Hz`

C

`200 Hz`

D

`150 Hz`

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The correct Answer is:
To solve the problem step by step, we will use the relationship between frequency and tension in a stretched string, as well as the concept of beat frequency. ### Step-by-Step Solution: 1. **Understanding the Relationship Between Frequency and Tension**: The frequency \( f \) of a stretched string is given by the formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] where \( T \) is the tension, \( L \) is the length of the string, and \( \mu \) is the linear mass density of the string. 2. **Change in Tension**: When the tension in one of the instruments is increased by 1%, the new tension \( T' \) can be expressed as: \[ T' = T + 0.01T = 1.01T \] 3. **Calculating the New Frequency**: The new frequency \( f' \) can be calculated using the new tension: \[ f' = \frac{1}{2L} \sqrt{\frac{T'}{\mu}} = \frac{1}{2L} \sqrt{\frac{1.01T}{\mu}} = \sqrt{1.01} \cdot f \] 4. **Beat Frequency**: The beat frequency \( f_b \) is given by the difference between the new frequency and the original frequency: \[ f_b = f' - f = \sqrt{1.01} \cdot f - f = (\sqrt{1.01} - 1)f \] 5. **Given Information**: We know that 3 beats are produced in 2 seconds, which means the beat frequency is: \[ f_b = \frac{3 \text{ beats}}{2 \text{ seconds}} = 1.5 \text{ beats/second} \] 6. **Setting Up the Equation**: Now we can set up the equation: \[ (\sqrt{1.01} - 1)f = 1.5 \] 7. **Solving for Frequency \( f \)**: Rearranging the equation gives: \[ f = \frac{1.5}{\sqrt{1.01} - 1} \] 8. **Calculating \( \sqrt{1.01} \)**: Approximating \( \sqrt{1.01} \): \[ \sqrt{1.01} \approx 1.005 \] Therefore: \[ \sqrt{1.01} - 1 \approx 0.005 \] 9. **Final Calculation**: Substituting this back into the frequency equation: \[ f \approx \frac{1.5}{0.005} = 300 \text{ Hz} \] ### Conclusion: The initial frequency of vibration of each wire is approximately **300 Hz**.

To solve the problem step by step, we will use the relationship between frequency and tension in a stretched string, as well as the concept of beat frequency. ### Step-by-Step Solution: 1. **Understanding the Relationship Between Frequency and Tension**: The frequency \( f \) of a stretched string is given by the formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} ...
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