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Standing waves are set up on two strings fixed at each end, as shown in the drawing. The two strings have the same tension and mass per unit length, but they differ in length by 0.57 cm. The waves on the shorter string propagate with a speed of 41.8 m/s, and the fundamental frequency of the shorter string is 225 Hz. Determine the beat frequency produced by the two standing waves.

Short Answer

Expert verified
The beat frequency is approximately 12.86 Hz.

Step by step solution

01

Understanding Standing Waves

In this exercise, we are dealing with standing waves on two strings. Both strings are fixed at both ends, have the same tension and mass per unit length, which implies the speed of the wave depends only on this tension and mass per unit length. The fundamental frequency of a vibrating string is given by the equation \( f = \frac{v}{2L} \), where \( v \) is the wave speed and \( L \) is the length of the string.
02

Calculate Length of Shorter String

Given the wave speed \( v = 41.8 \text{ m/s} \) and the fundamental frequency \( f = 225 \text{ Hz} \), we can rearrange the formula to find the length of the shorter string: \[ L = \frac{v}{2f} = \frac{41.8}{2 \times 225} \approx 0.0929 \text{ m} \].
03

Find Length of the Longer String

Since the shorter string is 0.57 cm shorter, the length of the longer string is \( L + 0.0057 \) meters. Therefore, the length \( L' \) of the longer string is \( 0.0929 + 0.0057 = 0.0986 \text{ m} \).
04

Calculate Frequency of Longer String

Using the same wave speed \( v = 41.8 \text{ m/s} \), the fundamental frequency for the longer string is \( f' = \frac{v}{2L'} = \frac{41.8}{2 \times 0.0986} \approx 212.14 \text{ Hz} \).
05

Calculate Beat Frequency

The beat frequency \( f_{beat} \) is the absolute difference between the two frequencies: \[ f_{beat} = |225 - 212.14| \approx 12.86 \text{ Hz} \].

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Key Concepts

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

Wave Speed and Its Importance
Wave speed is a crucial part of understanding the behavior of waves on strings. It represents how fast the wave travels along the medium, in this case, the string. Wave speed depends on two major factors: the tension in the string and the mass per unit length of the string. The greater the tension or the lighter the string, the faster the wave will travel. For the exercise we have, both strings have the same tension and mass per unit length, ensuring that the wave speed must be the same. This speed is given as 41.8 m/s, which we use to calculate other properties like frequency and string length. Understanding wave speed helps in determining how quickly vibrations from one spot will reach another spot along the string. It's also essential for predicting the frequencies at which standing waves will form.
Fundamental Frequency Calculation
The fundamental frequency, also known as the first harmonic, is the lowest frequency at which a string can vibrate. This is the frequency at which the wave pattern completes one full wave, fitting perfectly between the two fixed ends of the string. The formula to calculate this is given by: \[ f = \frac{v}{2L} \]where \( v \) is the wave speed and \( L \) is the length of the string. For the shorter string in the exercise, the fundamental frequency is stated as 225 Hz. Using the wave speed of 41.8 m/s, the length of the string can be computed, and it tells us how long the string must be to resonate at this frequency. Finding the fundamental frequency is critical in music and acoustics, as it forms the basis of the musical notes produced by stringed instruments.
Understanding Beat Frequency
Beat frequency occurs when two sound waves of slightly different frequencies interfere with each other. In the context of our exercise, it is what happens when two strings of similar, but not identical, lengths vibrate. The beat frequency is calculated by taking the absolute difference between the frequencies of the two waves. For our strings, the beat frequency \( f_{beat} \) is determined by: \[ f_{beat} = |f_1 - f_2| \]where \( f_1 \) and \( f_2 \) are the fundamental frequencies of the two strings. In this scenario, we calculate \( |225 \text{ Hz} - 212.14 \text{ Hz}| \approx 12.86 \text{ Hz} \). Beat frequency is often used in tuning musical instruments, where the beats can help identify small differences in pitch that are hard to detect by ear.
String Length Calculation and Its Methodology
When working with standing waves, particularly on strings, determining the length of the string is fundamental. From the exercise, we know the relationship between wave speed, frequency, and string length is vital. The formula \( L = \frac{v}{2f} \) helps us rearrange to find the length, knowing the other two values: wave speed \( v \) and frequency \( f \). In our example, the length of the shorter string is calculated using this formula, resulting in approximately 0.0929 meters. Since the longer string is 0.57 cm, or 0.0057 meters longer, its length is calculated by simply adding this to the shorter string鈥檚 length, ending up with 0.0986 meters. Understanding how to calculate string length allows one to adjust and design experiments or musical instruments for desired vibrations and tones.

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Most popular questions from this chapter

Divers working in underwater chambers at great depths must deal with the danger of nitrogen narcosis (the 鈥渂ends鈥), in which nitrogen dissolves into the blood at toxic levels. One way to avoid this danger is for divers to breathe a mixture containing only helium and oxygen. Helium, however, has the effect of giving the voice a high-pitched quality, like that of Donald Duck鈥檚 voice. To see why this occurs, assume for simplicity that the voice is generated by the vocal cords vibrating above a gas-filled cylindrical tube that is open only at one end. The quality of the voice depends on the harmonic frequencies generated by the tube; larger frequencies lead to higher-pitched voices. Consider two such tubes at \(20^{\circ} C\) . One is filled with air, in which the speed of sound is 343 \(m/s\) . The other is filled with helium, in which the speed of sound is \(1.00 \times 10^{3} m /s\) . To see the effect of helium on voice quality, calculate the ratio of the nth natural frequency of the helium-filled tube to the \(nth\) natural frequency of the air-filled tube.

An organ pipe is open at both ends. It is producing sound at its third harmonic, the frequency of which is 262 Hz. The speed of sound is 343 m/s. What is the length of the pipe?

A pipe open only at one end has a fundamental frequency of 256 Hz. A second pipe, initially identical to the first pipe, is shortened by cutting off a portion of the open end. Now, when both pipes vibrate at their fundamental frequencies, a beat frequency of 12 Hz is heard. How many centimeters were cut off the end of the second pipe? The speed of sound is 343 m/s.

The range of human hearing is roughly from twenty hertz to twenty kilohertz. Based on these limits and a value of 343 m/s for the speed of sound, what are the lengths of the longest and shortest pipes (open at both ends and producing sound at their fundamental frequencies) that you expect to find in a pipe organ?

Two carpenters are hammering at the same time, each at a different hammering frequency. The hammering frequency is the number of hammer blows per second. Every 4.6 s, both carpenters strike at the same instant, producing an effect very similar to a beat frequency. The first carpenter strikes a blow every 0.75 s. How many seconds elapse between the second carpenter鈥檚 blows if the second carpenter hammers (a) more rapidly than the first carpenter, and (b) less rapidly than the first carpenter?

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