In an acoustics experiment, a piano string with a mass of \(5.00 \mathrm{~g}\) and a length of \(70.0 \mathrm{~cm}\) is held under tension by running the string over a frictionless pulley and hanging a \(250 .-\mathrm{kg}\) weight from it. The whole system is placed in an elevator. a) What is the fundamental frequency of oscillation for the string when the elevator is at rest? b) With what acceleration and in what direction (up or down) should the elevator move for the string to produce the proper frequency of \(440 .\) Hz, corresponding to middle A?

Short Answer

Expert verified
Answer: The fundamental frequency of the string when the elevator is at rest is approximately 261.1 Hz. To produce a frequency of 440 Hz, the elevator needs to accelerate upwards with an acceleration of approximately 3.1244 m/s².

Step by step solution

01

Calculate tension in the string at rest

When the elevator is at rest, the tension in the string (T) is equal to the weight force acting on the suspended mass (m). To find the tension, we must first convert the mass (250g) to kilograms (0.25kg) and then multiply it by the acceleration due to gravity (g = 9.81 m/s²). T = m * g T = 0.25 kg * 9.81 m/s² T = 2.4525 N
02

Calculate linear density of the string

To find the linear density (μ) of the string, we must first convert the mass of the string (5.00g) to kilograms (0.005kg) and then divide it by the length of the string (70.0cm or 0.7m). μ = mass_of_string / length_of_string μ = 0.005 kg / 0.7 m μ = 0.007143 kg/m
03

Calculate the fundamental frequency at rest

Now, we can calculate the fundamental frequency (f) of the string using the formula: f = (1/2L) * sqrt(T/μ) Where L is the length of the string and T is the tension. f = (1 / (2 * 0.7 m)) * sqrt(2.4525 N / 0.007143 kg/m) f ≈ 261.1 Hz The fundamental frequency of oscillation for the string when the elevator is at rest is approximately 261.1 Hz.
04

Calculate the required tension for 440 Hz

To produce a frequency of 440 Hz, we must modify the tension in the string. Using the formula for fundamental frequency, we can solve for the required tension (T_new). f_new = 440 Hz T_new = (2L * f_new)^2 * μ T_new = (2 * 0.7 m * 440 Hz)^2 * 0.007143 kg/m T_new ≈ 4.0486 N
05

Calculate the required acceleration of the elevator

Next, we have to consider the force acting on the mass when the elevator is accelerating. The net force (F_net) acting on the mass is the difference between the weight force and the tension (T_new). Then we can use Newton's second law to find the acceleration (a) of the elevator. F_net = m * a = m * g - T_new a = (m * g - T_new) / m a = (0.25 kg * 9.81 m/s² - 4.0486 N) / 0.25 kg a ≈ 3.1244 m/s² Since the acceleration is positive, the elevator should move up for the string to produce the proper frequency of 440 Hz. In conclusion, when the elevator is at rest, the fundamental frequency of the string is approximately 261.1 Hz. To produce a frequency of 440 Hz, the elevator needs to accelerate upwards with an acceleration of approximately 3.1244 m/s².

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Key Concepts

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

Acoustics Experiment
Acoustics experiments often involve studying the properties of sound waves and how they interact with different materials and conditions. One key element in such experiments is the generation of sound through the vibration of a medium, like a string in a musical instrument. Understanding the frequency of a vibrating string is crucial, as it relates to the pitch of the note produced. In the context of our exercise, the fundamental frequency—that is, the lowest frequency at which the string naturally vibrates—is determined under static conditions and also when an external force (acceleration of the elevator) is applied. This illustrates how changes in the physical environment can affect sound production and frequency, offering a practical perspective into the physics of acoustics.
Tension in the String
Tension plays a pivotal role in determining the frequency of a string's vibration. Simply put, tension refers to the force that stretches the string, making it taut. The greater the tension, the faster the string vibrates, and thus the higher the pitch of the sound it produces. In an experiment like ours, tension is controlled by suspending a weight from the string, which stretches it by the force of gravity. However, if we alter the system—say, by moving the elevator up or down—we effectively change the tension in the string, either by adding to or subtracting from the force due to gravity, which in turn alters the fundamental frequency of the vibration.
Linear Density of the String
The linear density of a string is a measure of mass per unit length and is a key factor in the wave properties of the string, including its frequency. In our exercise, calculating the linear density requires knowledge of the string's total mass and its length. The linear density, indicated by the Greek letter μ (mu), determines how mass is distributed along the string and affects how quickly a wave can travel through it. A lower linear density means that less mass needs to be moved as the string vibrates, allowing for higher frequencies at the same tension. It's a principle that's utilized in designing musical instruments, as varying the linear density of strings allows for a wide range of pitches.
Newton's Second Law
Newton's second law is fundamental in classical mechanics and states that the acceleration of an object is directly proportional to the net force acting upon it, and inversely proportional to its mass. The formula is expressed as \( F = m \times a \), where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration. This principle is not just confined to solid objects, but also relates to waves in strings. In our example, suspending the weight from the string applies a force that we experience as tension. By moving the elevator, we apply additional force (or reduce it), thereby changing the tension and thus the acceleration of the mass, which, according to Newton's second law, must result in a modified force—and, as a result, a modified frequency of vibration for the string, as observed in the second part of our exercise.

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Most popular questions from this chapter

If two traveling waves have the same wavelength, frequency, and amplitude and are added appropriately, the result is a standing wave. Is it possible to combine two standing waves in some way to give a traveling wave?

A string with linear mass density \(\mu=0.0250 \mathrm{~kg} / \mathrm{m}\) under a tension of \(T=250 . \mathrm{N}\) is oriented in the \(x\) -direction. Two transverse waves of equal amplitude and with a phase angle of zero (at \(t=0\) ) but with different frequencies \((\omega=3000\). rad/s and \(\omega / 3=1000 . \mathrm{rad} / \mathrm{s}\) ) are created in the string by an oscillator located at \(x=0 .\) The resulting waves, which travel in the positive \(x\) -direction, are reflected at a distant point, so there is a similar pair of waves traveling in the negative \(x\) -direction. Find the values of \(x\) at which the first two nodes in the standing wave are produced by these four waves.

A small ball floats in the center of a circular pool that has a radius of \(5.00 \mathrm{~m}\). Three wave generators are placed at the edge of the pool, separated by \(120 .\). The first wave generator operates at a frequency of \(2.00 \mathrm{~Hz}\). The second wave generator operates at a frequency of \(3.00 \mathrm{~Hz}\). The third wave generator operates at a frequency of \(4.00 \mathrm{~Hz}\). If the speed of each water wave is \(5.00 \mathrm{~m} / \mathrm{s}\), and the amplitude of the waves is the same, sketch the height of the ball as a function of time from \(t=0\) to \(t=2.00 \mathrm{~s}\), assuming that the water surface is at zero height. Assume that all the wave generators impart a phase shift of zero. How would your answer change if one of the wave generators was moved to a different location at the edge of the pool?

A particular steel guitar string has mass per unit length of \(1.93 \mathrm{~g} / \mathrm{m}\). a) If the tension on this string is \(62.2 \mathrm{~N},\) what is the wave speed on the string? b) For the wave speed to be increased by \(1.0 \%\), how much should the tension be changed?

A sinusoidal wave on a string is described by the equation \(y=(0.100 \mathrm{~m}) \sin (0.75 x-40 t),\) where \(x\) and \(y\) are in meters and \(t\) is in seconds. If the linear mass density of the string is \(10 \mathrm{~g} / \mathrm{m}\), determine (a) the phase constant, (b) the phase of the wave at \(x=2.00 \mathrm{~cm}\) and \(t=0.100 \mathrm{~s}\) (c) the speed of the wave, (d) the wavelength, (e) the frequency, and (f) the power transmitted by the wave.

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