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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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Most popular questions from this chapter

A string of length \(70 \mathrm{~cm}\) is stretched between two rigid supports. The resonant frequency for this string is found to be \(420 \mathrm{~Hz}\) and \(315 \mathrm{~Hz}\). If there are no resonant frequencies between these two values, then what would be the minimum resonant frequency of this string ? (A) \(10.5 \mathrm{~Hz}\) (B) \(1.05 \mathrm{~Hz}\) (C) \(105 \mathrm{~Hz}\) (D) \(1050 \mathrm{~Hz}\)

The periodic time of a S.H.O. oscillating about a fixed point is $2 \mathrm{~s}$. After what time will the kinetic energy of the oscillator become \(25 \%\) of its total energy? (A) \(1 / 12 \mathrm{~s}\) (B) \(1 / 6 \mathrm{~s}\) (C) \(1 / 4 \mathrm{~s}\) (D) \(1 / 3 \mathrm{~s}\).

A spring is attached to the center of a frictionless horizontal turn table and at the other end a body of mass \(2 \mathrm{~kg}\) is attached. The length of the spring is \(35 \mathrm{~cm}\). Now when the turn table is rotated with an angular speed of \(10 \mathrm{rad} \mathrm{s}^{-1}\), the length of the spring becomes \(40 \mathrm{~cm}\) then the force constant of the spring is $\ldots \ldots \mathrm{N} / \mathrm{m}$. (A) \(1.2 \times 10^{3}\) (B) \(1.6 \times 10^{3}\) (C) \(2.2 \times 10^{3}\) (D) \(2.6 \times 10^{3}\)

A body having mass \(5 \mathrm{~g}\) is executing S.H.M. with an amplitude of \(0.3 \mathrm{~m}\). If the periodic time of the system is $(\pi / 10) \mathrm{s}\(, then the maximum force acting on body is \)\ldots \ldots \ldots \ldots$ (A) \(0.6 \mathrm{~N}\) (B) \(0.3 \mathrm{~N}\) (C) \(6 \mathrm{~N}\) (D) \(3 \mathrm{~N}\)

A rectangular block having mass \(\mathrm{m}\) and cross sectional area A is floating in a liquid having density \(\rho\). If this block in its equilibrium position is given a small vertical displacement, its starts oscillating with periodic time \(\mathrm{T}\). Then in this case \(\ldots \ldots\) (A) \(\mathrm{T} \propto(1 / \sqrt{\mathrm{m}})\) (B) \(T \propto \sqrt{\rho}\) (C) \(\mathrm{T} \propto(1 / \sqrt{\mathrm{A}})\) (D) \(\mathrm{T} \propto(1 / \sqrt{\rho})\)
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