/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 A stretched string of length \(L... [FREE SOLUTION] | 91Ó°ÊÓ

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A stretched string of length \(L\) is observed to vibrate in five equal segments when driven by a \(630-\mathrm{Hz}\) oscillator. What oscillator frequency will set up a standing wave so that the string vibrates in three segments?

Short Answer

Expert verified
Using the properties of standing waves and the given formula, it's possible to calculate the oscillator frequency of the second scenario. The specifics of the resulting frequency will depend on the length of the string, which was not given in the exercise.

Step by step solution

01

Identify known variables

From the exercise, the known values are the frequency of the oscillator as \(630-Hz\) and the segments of the string, which are five (n1=5). The string length remains the same 'L'.
02

Formulate a relationship

The relationship is given by the formula \(f1 = n1*v / 2*L\), where \(f1\) is the frequency, \(v\) velocity of wave, \(L\) the length of the string and \(n1\) the number of segments or modes. Because two situations with the same string (length, material etc.) are compared, it can be known that the velocity v remains constant.
03

Calculate v from known variables

Reformulate the formula to solve wave velocity \(v\) from the first case: \(v = 2*L*f1 / n1\)
04

Apply the velocity in the second case

Apply the calculated velocity into the same formula but for the other case, where the string vibrates in three segments (n2=3). Thus, the formula for the frequency in the second frequency \(f2\) will be: \(f2 = n2*v / 2*L\)
05

Compute the frequency for the second case

Since v, L and n are known now, using this formula, calculate \(f2\)

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Key Concepts

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

Harmonics in Strings
Understanding harmonics in strings is a fundamental part of studying waves and vibrations. Harmonics, or overtones, are vibrations that produce a standing wave pattern on a string. These are integral in music and physics, each corresponding to a distinct mode of vibration.

A string has different modes of vibration which can be prompted based on the frequency used to drive it. The simplest form of vibration occurs at the fundamental frequency, also called the first harmonic. This is the lowest frequency at which a string vibrates and is characterized by having two nodes at the ends and one antinode in the middle.

As the frequency increases, strings exhibit higher harmonics. The number of antinodes increases, and each standing wave pattern on the string is associated with a specific harmonic. For the exercise given, when the string vibrates in five equal segments, it is in its fifth harmonic. Altering the frequency to make the string vibrate in three segments will shift it to its third harmonic. The understanding of these harmonics is crucial in both the design of musical instruments and the study of wave properties.
Wave Velocity
Wave velocity, often represented by the symbol 'v', is an essential aspect of understanding wave behavior, especially in strings. It essentially denotes how fast a wave is propagating through a medium. In the context of strings, this medium is the string itself.

The velocity of a wave on a string is determined by the tension in the string and the mass per unit length of the string. However, in exercises like the one provided, if the string's tension and mass per unit length are constant, the wave velocity can be considered unchanged regardless of the frequency or harmonic considered.

The constant velocity of a wave in such cases allows for the application of the formula mentioned in the solution steps. By calculating wave velocity from a known frequency and number of segments, we can effectively reverse-engineer the problem to find a new frequency that would correspond to a different number of segments or harmonics while the characteristics of the string remain the same.
Frequency Calculation
Frequency calculation is a critical skill in physics, especially in wave mechanics and acoustics. The frequency of a wave is the number of times a repeating event occurs per unit time, such as the number of vibrations per second. It is measured in hertz (Hz).

In the context of the string exercise, we are interested in finding out what frequency is necessary to produce a certain harmonic on a string of constant length. Using the formula given in the solution, where the frequency is inversely proportional to the number of segments (modes of vibration), it becomes straightforward to determine the frequency for any harmonic.

Once the wave velocity has been calculated using a known frequency, that velocity can then be used to find the frequencies of other harmonics on the same string. This frequency calculation is vital not only for theoretical exercises but also for practical applications like tuning musical instruments or designing sound systems.

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Most popular questions from this chapter

How far, and in what direction, should a cellist move her finger to adjust a string's tone from an out-of-tune \(449 \mathrm{~Hz}\) to an in-tune \(440 \mathrm{~Hz}\) ? The string is \(68.0 \mathrm{~cm}\) long, and the finger is \(20.0 \mathrm{~cm}\) from the nut for the 449-Hz tone.

A standing wave is set up in a string of variable length and tension by a vibrator of variable frequency. Both ends of the string are fixed. When the vibrator has a frequency \(f_{A}\), in a string of length \(L_{A}\) and under tension \(T_{A}, n_{A}\) antinodes are set up in the string. (a) Write an expression for the frequency \(f_{A}\) of a standing wave in terms of the number \(n_{A}\), length \(L_{A}\), tension \(T_{A}\), and linear density \(\mu_{A^{*}}\) (b) If the length of the string is doubled to \(L_{B}=2 L_{A}\), what frequency \(f_{B}\) (written as a multiple of \(f_{A}\) ) will result in the same number of antinodes? Assume the tension and linear density are unchanged. Hint: Make a ratio of expressions for \(f_{B}\) and \(f_{A}\). (c) If the frequency and length are held constant, what tension \(T_{B}\) will produce \(n_{A}+1\) antinodes? (d) If the frequency is tripled and the length of the string is halved, by what factor should the tension be changed so that twice as many antinodes are produced?

By proper excitation, it is possible to produce both longitudinal and transverse waves in a long metal rod. In a particular case, the rod is \(150 \mathrm{~cm}\) long and \(0.200 \mathrm{~cm}\) in radius and has a mass of \(50.9 \mathrm{~g}\). Young's modulus for the material is \(6.80 \times 10^{10} \mathrm{~Pa}\). Determine the required tension in the rod so that the ratio of the speed of longitudinal waves to the speed of transverse waves is 8 .

A student holds a tuning fork oscillating at \(256 \mathrm{~Hz}\). He walks toward a wall at a constant speed of \(1.33 \mathrm{~m} / \mathrm{s}\). (a) What beat frequency does he observe between the tuning fork and its echo? (b) How fast must he walk away from the wall to observe a beat frequency of \(5.00 \mathrm{~Hz}\) ?

Two pieces of steel wire with identical cross sections have lengths of \(L\) and \(2 L\). The wires are each fixed at both ends and stretched so that the tension in the longer wire is four times greater than in the shorter wire. If the fundamental frequency in the shorter wire is \(60 \mathrm{~Hz}\), what is the frequency of the second harmonic in the longer wire?

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