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Chapter 21: Problem 29

Two strings are adjusted to vibrate at exactly \(200 \mathrm{Hz}\). Then the tension in one string is increased slightly. Afterward, three beats per second are heard when the strings vibrate at the same time. What is the new frequency of the string that was tightened?

Short Answer

Expert verified
The new frequency of the string whose tension was increased is 203 Hz.

Step by step solution

01

Understand and apply the beat frequency formula

According to the beat frequency formula, the beat frequency is equal to the absolute difference of the frequencies of the two sound sources. This would mean: \(Beat Frequency = |f_1 - f_2|\), where \(f_1\) and \(f_2\) are the frequencies of the two sources. In this example, the original frequency of both strings before the tension was increased, \(f_1 = f_2 = 200 Hz\), and the beat frequency after the tension was increased is given as 3 Hz.
02

Calculate the new frequency of the tightened string

We know that the beat frequency is obtained from the absolute difference between the frequencies of the two strings. The new frequency of the tightened string, \(f_{new}\), can therefore be calculated from the given beat frequency, \(f_{beat}\) = |200 - \(f_{new}\)| = 3 Hz. Either \(f_{new}\) = 200 + 3 = 203 Hz or \(f_{new}\) = 200 - 3 = 197 Hz; Since we are told that the tension on the string is increased, the frequency \(f_{new}\) also increases, so only the first solution makes sense. Therefore \(f_{new}\) = 203 Hz.

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Most popular questions from this chapter

A \(50-\mathrm{cm}\) -long wire with a mass of \(1.0 \mathrm{g}\) and a tension of \(440 \mathrm{N}\) passes across the open end of an open-closed tube of air. The wire, which is fixed at both ends, is bowed at the center so as to vibrate at its fundamental frequency and generate a sound wave. Then the tube length is adjusted until the fundamental frequency of the tube is heard. What is the length of the tube? Assume \(v_{\text {sound }}=340 \mathrm{m} / \mathrm{s}\).

a. The frequency of a standing wave on a string is \(f\) when the string's tension is \(T\). If the tension is changed by the small amount \(\Delta T\), without changing the length, show that the frequency changes by an amount \(\Delta f\) such that 1 $$ \frac{\Delta f}{f}=\frac{1}{2} \frac{\Delta T}{T} $$ b. Two identical strings vibrate at \(500 \mathrm{Hz}\) when stretched with the same tension. What percentage increase in the tension of one of the strings will cause five beats per second when both strings vibrate simultancously?

A carbon dioxide laser is an infrared laser. A \(\mathrm{CO}_{2}\) laser with a cavity length of 53.00 cm oscillates in the \(m=100,000\) mode. What are the wavelength and frequency of the laser beam?

Two loudspeakers emit sound waves along the \(x\) -axis. Speaker 2 is 2.0 m behind speaker \(1 .\) Both loudspeakers are connected to the same signal generator, which is oscillating at \(340 \mathrm{Hz}\), but the wire to speaker 1 passes through a box that delays the signal by \(1.47 \mathrm{ms} .\) Is the interference along the \(x\) -axis maximum constructive interference, perfect destructive interference, or something in between? Assume \(v_{\text {sund }}=340 \mathrm{m} / \mathrm{s}\).

A string of length \(L\) vibrates at its fundamental frequency. The amplitude at a point \(\frac{1}{4} L\) from one end is \(2.0 \mathrm{cm} .\) What is the amplitude of each of the traveling waves that form this standing wave?
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