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Two identical strings on different cellos are tuned to the 440 -Hz A note. The peg holding one of the strings slips, so its tension is decreased by \(1.5 \% .\) What is the beat frequency heard when the strings are then played together?

Short Answer

Expert verified
The beat frequency heard is 3.3 Hz.

Step by step solution

01

Identify Known Values

The original frequency of both strings is 440 Hz. The tension of one string is reduced by 1.5%.
02

Relate Frequency to Tension

The frequency of a vibrating string is proportional to the square root of its tension. This means if tension changes by a certain percentage, the frequency changes by half that percentage. So, a 1.5% decrease in tension results in approximately a 0.75% decrease in frequency.
03

Calculate the New Frequency

Calculate the new frequency of the slipped string. The frequency decreases by 0.75%: \[ f_{new} = 440 \text{ Hz} \times (1 - 0.0075) \approx 436.7 \text{ Hz} \]
04

Determine the Beat Frequency

The beat frequency is the absolute difference between the two frequencies: \[ f_{beat} = |440 \text{ Hz} - 436.7 \text{ Hz}| = 3.3 \text{ Hz} \].

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Key Concepts

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

Vibrating Strings
String instruments, like cellos, create sound through the vibration of strings. When a string vibrates, it disturbs the surrounding air molecules, initiating waves that travel to our ears as sound. Each string on an instrument is designed to vibrate at a specific frequency when played, corresponding to a musical note. The frequency, which is the number of vibrations per second, determines the pitch of the sound produced. For example, a frequency of 440 Hz corresponds to the note A in music standard tuning. If a string is fixed at both ends, as in instruments, it vibrates in a series of modes, principally determined by its tension, length, and mass. Players can change the pitch by altering any of these factors.
Tension and Frequency Relationship
The relationship between the tension in a string and its frequency is crucial in string instruments. The frequency at which a string vibrates is directly proportional to the square root of its tension. This means that increasing the tension in a string will increase its frequency, raising the pitch of the note. Conversely, decreasing the tension will lower the frequency, resulting in a deeper pitch.Mathematically, this can be expressed as:\[f \ ext{ 鈭 } \ \sqrt{T}\]where \( f \) is frequency and \( T \) is tension. This principle allows musicians to fine-tune their instruments by adjusting the tension, hence achieving the desired pitch for each string.
Frequency Change Due to Tension
When tension in a string is altered, the frequency changes, but not in a linear way. Instead, the frequency changes proportional to the square root of the tension change. If the tension decreases by a certain percentage, the frequency decreases by approximately half that percentage. For example, in our cello string scenario, a tension decrease of 1.5% results in a frequency decrease of about 0.75%.We can calculate this by noting the percentage change in tension as \( rac{ ext{Percentage Change in Tension}}{2} \). With that understanding, we can predict how the pitch will drop when the tension is reduced, using calculations to find the new frequency of the string.
Calculating Beat Frequency
Beat frequency occurs when two strings are played together, but are slightly out of tune with one another. This happens when their frequencies differ, leading to periodic variations in sound intensity, which we perceive as beats. The number of beats per second is the beat frequency.To calculate the beat frequency, you need to find the absolute difference between the two strings' frequencies. Using the cello example, one string, undeviated, still vibrates at 440 Hz, while the other, with decreased tension, vibrates at approximately 436.7 Hz.The beat frequency calculated is:\[f_{beat} = |440 ext{ Hz} - 436.7 ext{ Hz}| = 3.3 ext{ Hz}\]This tells us that when these slightly different pitches are played together, we hear a beating effect of 3.3 cycles per second.

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Most popular questions from this chapter

How fast, in kilometers per hour, must a sound source be moving toward you to make the observed frequency \(5.0 \%\) greater than the true frequency? (Assume that the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\).)

A whale is swimming at a steady speed either directly at or directly away from an underwater cliff (you don't know which). When the whale is \(300 \mathrm{~m}\) from the cliff, it emits a sound and it hears the echo \(0.399 \mathrm{~s}\) later. (a) Which way is the whale traveling: (i) toward the cliff or (ii) away from the cliff? Explain your reasoning. (b) How fast is the whale traveling? (c) If the emitted sound has a frequency of \(12.1 \mathrm{kHz}\), by how much has the frequency changed by the time the whale hears the echo?

While standing near a railroad crossing, you hear a train horn. The frequency emitted by the horn is \(400 \mathrm{~Hz}\). If the train is traveling at \(90.0 \mathrm{~km} / \mathrm{h}\) and the air temperature is \(25^{\circ} \mathrm{C},\) what is the frequency you hear (a) when the train is approaching and (b) after it has passed?

An unstretchable steel string is used to replace a broken violin string. A length of \(5.00 \mathrm{~m}\) of this string has a mass of \(25.0 \mathrm{~g}\). When in place, the new string will be \(30.0 \mathrm{~cm}\) long and oscillate at \(256 \mathrm{~Hz}\) in its fundamental mode. (a) After it is in place, what tension must the new string be placed under? (b) Assuming normal room temperature, what is the wavelength of the sound emitted by the new string in its fundamental mode? (c) If you wanted to decrease this sound wavelength by \(5.00 \%,\) what would you do to the tension: increase or decrease it? Explain. (d) Determine the required tension in part (c).

A bat moving at \(15.0 \mathrm{~m} / \mathrm{s}\) emits a high-frequency sound as it approaches a wall that is \(25.0 \mathrm{~m}\) away. Assuming that the bat continues straight toward the wall, how far away is it when it receives the echo? (Assume the air temperature in the cave to be \(\left.0^{\circ} \mathrm{C} .\right)\)

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