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Chapter 10: Problem 1490

The frequency of tuning fork \(\mathrm{A}\) is \(2 \%\) more than the frequency of a standard fork. Frequency of tuning fork \(B\) is \(3 \%\) less than the frequency of the standard fork. If 6 beats per second are heard when the two forks \(\mathrm{A}\) and \(\mathrm{B}\) are excited, then frequency of \(A\) is \(\ldots \ldots \ldots . H z\) (A) 120 (B) \(122.4\) (C) \(116.4\) (D) 130

Short Answer

Expert verified
The frequency of tuning fork A is \(122.4\, Hz\), which corresponds to option (B).

Step by step solution

01

Setting up the equations

Let the standard frequency be S. According to the problem, frequency of A is 2% more than S, so: \(A = S + 0.02S = 1.02S\) Similarly, frequency of B is 3% less than S, so: \(B = S - 0.03S = 0.97S\) We also know that the absolute difference between the frequencies of A and B is 6 Hz. \(|A - B| = 6\)
02

Solving for the standard frequency (S)

We know that \(A = 1.02S\) and \(B = 0.97S\). Substitute these into the equation \(|A - B| = 6\), we get: \(|1.02S - 0.97S| = 6\) Solve for S: \(|0.05S| = 6\) \(S = \frac{6}{0.05}\) \(S = 120\)
03

Find the frequency of A

We now know the standard frequency is 120 Hz. To find the frequency of A, we substitute the value of S into the equation: \(A = 1.02S\) \(A = 1.02 \times 120\) \(A = 122.4\, Hz\) Therefore, the frequency of tuning fork A is 122.4 Hz, which corresponds to option (B).

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Most popular questions from this chapter

A tuning fork arrangement produces 4 beats/second with one fork of frequency \(288 \mathrm{~Hz}\). A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is $\ldots \ldots \ldots . \mathrm{Hz}$. (A) 286 (B) 292 (C) 294 (D) 288

A simple pendulum having length \(\ell\) is suspended at the roof of a train moving with constant acceleration 'a' along horizontal direction. The periodic time of this pendulum is.... (A) \(\mathrm{T}=2 \pi \sqrt{(\ell / \mathrm{g})}\) (B) \(\mathrm{T}=2 \pi \sqrt{\\{\ell /(\mathrm{g}+\mathrm{a})\\}}\) (C) \(\mathrm{T}=2 \pi \sqrt{\\{\ell /(\mathrm{g}-\mathrm{a})\\}}\) (D) $\left.\mathrm{T}=2 \pi \sqrt{\\{\ell} /\left(\mathrm{g}^{2}+\mathrm{a}^{2}\right)\right\\}$

Twenty four tuning forks are arranged in such a way that each fork produces 6 beats/s with the preceding fork. If the frequency of the last tuning fork is double than the first fork, then the frequency of the second tuning fork is \(\ldots \ldots\) (A) 132 (B) 138 (C) 276 (D) 144

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) \(2.6 \times 10^{3}\)

One end of a mass less spring having force constant \(\mathrm{k}\) and length \(50 \mathrm{~cm}\) is attached at the upper end of a plane inclined at an angle \(e=30^{\circ} .\) When a body of mass \(m=1.5 \mathrm{~kg}\) is attached at the lower end of the spring, the length of the spring increases by $2.5 \mathrm{~cm}$. Now, if the mass is displaced by a small amount and released, the amplitude of the resultant oscillation is ......... (A) \((\pi / 7)\) (B) \((2 \pi / 7)\) (C) \((\pi / 5)\) (D) \((2 \pi / 5)\)
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