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Chapter 16: Problem 80

A fork of unknown frequency gives four beats \(\mathrm{s}^{-1}\) when sounded with another of frequency 256 . The fork is now loaded with a piece of wax and again four beats \(\mathrm{s}^{-1}\) are heard. Then the frequency of the unknown fork is (a) \(256 \mathrm{~Hz}\) (b) \(252 \mathrm{~Hz}\) (c) \(264 \mathrm{~Hz}\) (d) \(260 \mathrm{~Hz}\)

Short Answer

Expert verified
The frequency of the unknown fork is 260 Hz.

Step by step solution

01

Understanding the Concept of Beats

When two sound waves of slightly different frequencies interfere with each other, they create a phenomenon called 'beats'. The beat frequency is the absolute difference between the two frequencies, given by the formula: \( |f_1 - f_2| \). Here, \( f_1 \) is the frequency of the known tuning fork, and \( f_2 \) is the unknown frequency.
02

Identifying Initial Conditions

The problem states that the unknown fork, when sounded together with a fork of frequency 256 Hz, produces 4 beats per second. This implies that the absolute difference \(|f_2 - 256|\) is equal to 4 beats per second.
03

Setting Up the Initial Equation

From the beat frequency concept, we write the equation: \(|f_2 - 256| = 4\). This gives us two possible frequencies for the unknown fork: \(f_2 = 256 + 4 = 260\,\text{Hz}\) or \(f_2 = 256 - 4 = 252\,\text{Hz}\).
04

Considering the Wax Loading Effect

Adding wax to a tuning fork typically lowers its frequency. Since adding wax still results in 4 beats per second, the frequency shift should still allow the condition \(|f_2 - 256| = 4\) to hold true with the frequency lowered.
05

Determining the Correct Frequency

For the frequency to be 4 Hz less than 256 Hz even after adding wax, the original frequency must be \(260\,\text{Hz}\), because adding wax reduces the frequency 鈥� thus initially higher frequency results, when reduced, continue the same beat condition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Beat Frequency
When two sound waves of close yet different frequencies encounter each other, they create an interesting auditory pattern called 'beats'. This occurs as the waves periodically reinforce and cancel one another. You can think of it like when you hear sudden loudness and quietness in a rhythmic manner. The beat frequency determines the rate at which these variations in loudness happen.

To compute this beat frequency, we use a simple formula: \( |f_1 - f_2| \). Here, \( f_1 \) and \( f_2 \) represent the two varying frequencies. The absolute value is crucial here because it provides a non-negative result, giving you the number of beats per second without directional indication.
  • This beat effect is especially useful for detecting small differences in frequency, such as when tuning musical instruments.
  • The phenomena is observable when the frequency difference is less than 10 Hz, making it very efficient for precise auditory checks.
Understanding beat frequency is critical in this exercise, as it helps deduce the unknown frequency based on known parameters.
Tuning Fork Frequency
Tuning forks are devices that vibrate at a set frequency when struck, making them valuable for matching specific pitches. Each tuning fork is defined by its frequency, measured in Hertz (Hz). In the exercise, one of the forks has a known frequency of 256 Hz.

Our goal is to find the frequency of another tuning fork that, when played with the 256 Hz fork, results in 4 beats per second. Here are the crucial considerations:
  • The formula \(|f_2 - 256| = 4\) reveals the possible values for the unknown frequency, \(f_2\), as either 252 Hz or 260 Hz.
  • These calculations show that the unknown fork could be either slightly higher or slightly lower than the known frequency, resulting in the same beat frequency.
This part of the challenge involves identifying possible frequency changes when two forks are engaged, highlighting the suspense in determining the correct frequency.
Effects of Wax on Frequency
Adding a piece of wax to a tuning fork is analogous to slowing down a race. The wax increases the mass of the fork's prongs, which typically causes the frequency to decrease. This is often visible in musical contexts where dampening alters pitch.

In this exercise, after adding wax to the unknown fork, 4 beats per second are still heard with a fork of frequency 256 Hz.
  • The presence of wax suggests that the fork originally had a higher frequency, at 260 Hz, because the added wax would lower its frequency, potentially towards 256 Hz.
  • Since the beat frequency remains unchanged, we deduce the fork's initial frequency must allow for this predictable reduction to continue matching the beat condition, hinting the original was 260 Hz.
Understanding this shift is key to solving the problem, as we use the behavior of wax to uncover the correct starting point for the fork's frequency.

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Most popular questions from this chapter

Which of the following statements are true for wave motion? (a) Mechanical transverse waves can propagate through all mediums. (b) Longitudinal waves can propagate through solids only. (c) Mechanical transverse waves can propagate through solids only. (d) Longitudinal waves can propagate through vacuum.

A wire of density \(9 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) is stretched between two clamps \(1 \mathrm{~m}\) apart and is subjected to an extension of \(4.9 \times 10^{-4} \mathrm{~cm} .\) The lowest frequency of transverse vibration in the wire is ( \(Y=9 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\) ) (a) \(40 \mathrm{~Hz}\) (b) \(35 \mathrm{~Hz}\) (c) \(30 \mathrm{~Hz}\) (d) \(25 \mathrm{~Hz}\)

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of \(45 \mathrm{~Hz}\). The mass of the wire is \(3.5 \times 10^{-2} \mathrm{~kg}\) and its linear mass density is \(4.0 \times 10^{-2} \mathrm{~kg} / \mathrm{m}\). What is (i) the speed of a transverse wave on the string and (ii) the tension in the string? (a) (i) \(80 \mathrm{~m} / \mathrm{s}\) (ii) \(250 \mathrm{~N}\) (b) (i) \(88 \mathrm{~m} / \mathrm{s}\) (ii) \(208 \mathrm{~N}\) (c) (i) \(90 \mathrm{~m} / \mathrm{s}\) (ii) \(249 \mathrm{~N}\) (d) (i) \(78.75 \mathrm{~m} / \mathrm{s}\) (ii) \(248 \mathrm{~N}\)

Two radio stations broadcast their programmes at the same amplitude \(A\) and at slightly different frequenies \(\omega_{1}\) and \(\omega_{2}\) respectively where \(\omega_{2}-\omega_{1}=10^{3} \mathrm{~Hz} . \mathrm{A}\) detector is receiving signals from the two stations simultaneously. It can only detect signals of intensity \(>2 A^{2}\). The time interval between successive maxima of the intensity of the signal received by the detector is (a) \(10^{3} \mathrm{~s}\) (b) \(10^{-3} \mathrm{~s}\) (c) \(10^{-4} \mathrm{~s}\) (d) \(10^{4} s\)

The frequency of a sonometer wire is \(100 \mathrm{~Hz}\). When the weights producing the tension are completely immersed in water, the frequency becomes \(80 \mathrm{~Hz}\) and on immersing the weights in a certain liquid, the frequency becomes \(60 \mathrm{~Hz}\). The specific gravity of the liquid is (a) \(1.42\) (b) \(1.77\) (c) \(1.21\) (d) \(1.82\)
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