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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

Two loudspeakers emit sound waves of the same frequency along the \(x\) -axis. The amplitude of each wave is \(a\). The sound intensity is minimum when speaker 2 is \(10 \mathrm{cm}\) behind speaker \(1 .\) The intensity increases as speaker 2 is moved forward and first reaches maximum, with amplitude \(2 a\), when it is \(30 \mathrm{cm}\) in front of speaker \(1 .\) What is a. The wavelength of the sound? b. The phase difference between the two loudspeakers? c. The amplitude of the sound (as a multiple of \(a\) ) if the speak- ers are placed side by side?

As the captain of the scientific team sent to Planet. Physics, one of your tasks is to measure \(g\). You have a long, thin wire labeled \(1.00 \mathrm{g} / \mathrm{m}\) and a \(1.25 \mathrm{kg}\) weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the \(1.25 \mathrm{kg}\) weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is \(100 \mathrm{Hz}\). Next, with the \(1.25 \mathrm{kg}\) weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 314 s to complete 100 oscillations. Pulling out your trusty calculator, you get to work. What value of \(g\) will you report back to headquarters?

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?

An open-open organ pipe is \(78.0 \mathrm{cm}\) long. An open-closed pipe has a fundamental frequency equal to the third harmonic of the open-open pipe. How long is the open-closed pipe?

A very thin oil film \((n=1.25)\) floats on water \((n=1.33)\) What is the thinnest film that produces a strong reflection for green light with a wavelength of \(500 \mathrm{nm} ?\)
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