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Chapter 10: Problem 1483

A tuning fork of frequency \(200 \mathrm{~Hz}\) is in unison with a sonometer wire. The number of beats heard per second when the tension is increased by $1 \%\( is \)\ldots \ldots \ldots .$ (A) 1 (B) 2 (C) 4 (D) \(0.5\)

Short Answer

Expert verified
The number of beats heard per second when the tension is increased by 1% is 1 (option A).

Step by step solution

01

Find the initial frequency of the sonometer wire

Since the tuning fork and sonometer wire are initially in unison, their frequencies are the same. Therefore, the initial frequency of the sonometer wire, \(f_1\), is 200 Hz.
02

Determine the new tension of the sonometer wire

We know that the tension in the wire has been increased by 1%. Let the initial tension be \(T\). Therefore, the new tension, \(T'\), can be found using the following relation: \[T' = T + 0.01T\]
03

Calculate the change in frequency due to the change in tension

The frequency of a sonometer wire is related to its tension by the following equation: \(f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\) where \(f\) is the frequency, \(L\) is the length of the wire, and \(\mu\) is the linear density of the wire. When the tension changes, the new frequency, \(f'\), can be found using the same equation: \(f' = \frac{1}{2L}\sqrt{\frac{T'}{\mu}}\)
04

Solve for the new frequency, \(f'\)

We need to find the ratio of the new frequency to the old frequency to find the new frequency. Using the equation for \(f\) and \(f'\) above and dividing the equations, we get: \(\frac{f'}{f} = \frac{\sqrt{T'}}{\sqrt{T}}\) Since \(f = 200 \, Hz\), we can find the new frequency \(f'\) using the new tension, \(T'\): \(f' = 200 \, Hz \cdot \frac{\sqrt{T'}}{\sqrt{T}} = 200 \, Hz \cdot \sqrt{\frac{T'}{T}}\) Substitute the equation for the new tension, \(T'\), from Step 2: \(f' = 200 \, Hz \cdot \sqrt{\frac{T + 0.01T}{T}} = 200 \, Hz \cdot \sqrt{1 + 0.01}\)
05

Calculate the number of beats heard per second

The beat frequency can be found by taking the difference between the new frequency, \(f'\), and the initial frequency, \(f\): Beat frequency = \(f' - f\) Using the equation for \(f'\) found in Step 4: Beat frequency = \(200 \, Hz \cdot \sqrt{1 + 0.01} - 200 \, Hz\) After evaluating the expression, we find that the beat frequency is approximately 1 Hz. Therefore, the number of beats heard per second when the tension is increased by 1% is 1 (option A).

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