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Chapter 10: Problem 1450

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

Short Answer

Expert verified
The frequency of the second tuning fork is 144 Hz.

Step by step solution

01

Understanding the concept of beats/s between two tuning forks

Beat frequency is the difference in frequencies between two waves meeting at the same time and place. In this case, each tuning fork produces 6 beats/s with the preceding fork, which means the difference in frequency between consecutive forks is 6 Hz.
02

Create an equation representing the relationship between the frequencies of the forks

Let f1 represent the frequency of the first fork. Then the frequency of the second fork can be expressed as f1 + 6, the third fork as f1 + 12, and so on. So, the frequency of the nth fork can be represented as f1 + 6(n - 1), where n is the position of the fork in the sequence. We are given that the frequency of the last fork, which is the 24th fork, is double the frequency of the first fork. So, we can write this as f1 + 6(24 - 1) = 2f1.
03

Solve for the frequency of the first fork

Now, we can solve the equation f1 + 6(24 - 1) = 2f1 to find the frequency of the first fork, f1: f1 + 6(23) = 2f1 f1 + 138 = 2f1 138 = f1 So, the frequency of the first fork is 138 Hz.
04

Find the frequency of the second fork

Now that we know the frequency of the first fork, we can determine the frequency of the second fork using the difference given at the start: Frequency of the second fork = f1 + 6 = 138 + 6 = 144 Hz Thus, the frequency of the second tuning fork is 144 Hz. The correct answer is (D) 144.

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Most popular questions from this chapter

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

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