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A heavy piece of hanging sculpture is suspended by a \(90-\mathrm{cm}-\) long, \(5.0 \mathrm{g}\) steel wire. When the wind blows hard, the wire hums at its fundamental frequency of \(80 \mathrm{Hz}\). What is the mass of the sculpture?

Short Answer

Expert verified
The mass of the sculpture is approximately \(37.1 \mathrm{kg}\).

Step by step solution

01

Calculating the tension

The tension in the wire when the sculpture is hanging equals to the weight of the sculpture. This tension also gives the wire its fundamental frequency. Hence, let's first calculate the tension which can be obtained using the formula of fundamental frequency for a string, which is, \(f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\), where \(L\) is the length, \(f\) is the fundamental frequency, \(T\) is the tension and \(\mu\) is the linear density. Linear density or mass per unit length, of the wire can be calculated as \(\mu = \frac{m_{\text{wire}}}{L}\) where \(m_{\text{wire}}\) is mass of the wire. Substituting values, \(\mu = \frac{5.0 \times 10^{-3} \mathrm{kg}}{90 \times 10^{-2} \mathrm{m}} = 5.56 \times 10^{-5} \mathrm{kg/m}\).
02

Isolate Tension in the Frequency Formula

Re-organizing the fundamental frequency formula to solve for \(T\) we get, \(T = \mu (2Lf)^2\). Substituting the values into the rearranged formula, we get \(T = 5.56 \times 10^{-5} \mathrm{kg/m} \times (2 \times 90 \times 10^{-2} \mathrm{m} \times 80 \mathrm{s}^{-1})^2 = 363.54 \mathrm{N}\).
03

Determine the sculpture's mass

The tension found in Step 2 is equal to the weight of the sculpture, which is the mass of the sculpture times the acceleration due to gravity. By rearranging the formula for weight (\(mg = T\)), we can find the mass of the sculpture: \(m_{\text{sculpture}} = \frac{T}{g}\). Substituting the values, \(m_{\text{sculpture}} = \frac{363.54 \mathrm{N}}{9.8 \mathrm{m/s}^2} = 37.1 \mathrm{kg}\)

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Key Concepts

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

Tension in a String
Imagine a guitar string fastened at both ends. When you pluck it, the string vibrates, producing sound. The tension in the string is a crucial factor that affects the speed of the wave traveling through it, and thereby, the pitch of the note it produces. More precisely, tension refers to the force exerted along the string's length.

In the context of our exercise, the tension in the wire holding the sculpture is essential for calculating the wire's fundamental frequency. The sculpture's weight creates a force due to gravity that pulls downwards. This force is countered by an upward tension force in the wire. Thus, tension equals the weight of the sculpture, which is a product of its mass and acceleration due to gravity.

Furthermore, the fundamental frequency of a string is directly related to the tension: as the tension increases, so does the frequency. For a given linear density and length, the tension can be computed knowing the fundamental frequency, which enables us to dedote the mass of an object suspended from the string, such as the sculpture in our exercise.
Linear Density
Linear density is a property that represents mass per unit length of a material or object, denoted by the symbol \(\mu\). In our study of strings and wires, it's a critical value because it affects how the string vibrates when disturbed. Linear density can be calculated by dividing the mass of the string or wire by its total length.

In our exercise, we work out the linear density of the steel wire to further understand how the tension and the fundamental frequency are related. With a lighter string, for the same tension, the frequency would be higher, producing a higher pitch. Similarly, a heavier string at the same tension vibrates slower and thus has a lower pitch. This concept is pivotal when solving problems in musical acoustics and material sciences where understanding the relationship between physical properties and behavior is necessary.
Standing Waves
Standing waves are patterns of vibration that appear to be stationary, meaning parts of the wave are not moving (nodes), and parts of the wave are moving with maximum amplitude (antinodes). When you pluck a stretched string, standing waves form as the wave reflects off the fixed ends of the string and interferes with incoming waves.

A string can support many different standing wave patterns, but the simplest and lowest frequency pattern is known as the fundamental frequency or first harmonic. This is what our exercise referred to. The fundamental frequency is determined by the string's length, tension, and linear density. Each of these parameters combine to define how the string vibrates and thus how it converts mechanical vibrations into waves of pressure that we perceive as sound. Underscoring the significance of this concept, musicians tune instruments by adjusting string tension to reach exact standing wave patterns that produce the desired pitches.

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Most popular questions from this chapter

Ultrasound has many medical applications, one of which is to monitor fetal heartbeats by reflecting ultrasound off a fetus in the womb. a. Consider an object moving at speed \(v_{\mathrm{o}}\) toward an at-rest source that is emitting sound waves of frequency \(f_{0} .\) Show that the reflected wave (i.c., the echo) that returns to the source has a Doppler- shifted frequency $$f_{\text {echo }}=\left(\frac{v+v_{\mathrm{o}}}{v-v_{\mathrm{o}}}\right) f_{0}$$ where \(v\) is the speed of sound in the medium. b. Suppose the object's speed is much less than the wave speed: \(v_{\mathrm{o}} \ll v .\) Then \(f_{\text {echo }} \approx f_{0},\) and a microphone that is sensitive to these frequencies will detect a beat frequency if it listens to \(f_{0}\) and \(f_{\text {echo }}\) simultaneously. Use the binomial approximation and other appropriate approximations to show that the beat frequency is \(f_{\text {beat }} \approx\left(2 v_{0} / v\right) f_{0}\) c. The reflection of \(2.40 \mathrm{MHz}\) ultrasound waves from the sur- face of a fetus's beating heart is combined with the \(2.40 \mathrm{MHz}\) wave to produce a beat frequency that reaches a maximum of \(65 \mathrm{~Hz}\). What is the maximum speed of the surface of the heart? The speed of ultrasound waves within the body is \(1540 \mathrm{~m} / \mathrm{s}\) d. Suppose the surface of the heart moves in simple harmonic motion at 90 beats/min. What is the amplitude in \(\mathrm{mm}\) of the heartbeat?

Two loudspeakers face each other from opposite walls of a room. Both are playing exactly the same frequency, thus setting up a standing wave with distance \(\lambda / 2\) between antinodes. Assume that \(\lambda\) is much less than the room width, so there are many antinodes. a. Yvette starts at one speaker and runs toward the other at speed \(v_{Y} .\) As the does so, she hears a loud-soft-loud modulation of the sound intensity. From your perspective, as you sit at rest in the room, Yvette is running through the nodes and antinodes of the standing wave. Find an expression for the number of sound maxima she hears per second. b. From Yvette's perspective, the two sound waves are Doppler shifted. They're not the same frequency, so they don't create a standing wave. Instead, she hears a loud-soft-loud modulation of the sound intensity because of beats. Find an expression for the beat frequency that Yvette hears. c. Are your answers to parts a and b the same or different? Should they be the same or different?

Two sinusoidal waves with equal wavelengths travel along a string in opposite directions at \(3.0 \mathrm{m} / \mathrm{s}\). The time between two successive instants when the antinodes are at maximum height is \(0.25 \mathrm{s} .\) What is the wavelength?

Two in-phase speakers \(2.0 \mathrm{m}\) apart in a plane are emitting. \(1800 \mathrm{Hz}\) sound waves into a room where the speed of sound is \(340 \mathrm{m} / \mathrm{s} .\) Is the point \(4.0 \mathrm{m}\) in front of one of the speakers, perpendicular to the plane of the speakers, a point of maximum constructive interference, perfect destructive interference, or something in between?

An old mining tunnel disappears into a hillside. You would like to know how long the tunnel is, but it's too dangerous to go inside. Recalling your recent physics class, you decide to try setting up standing-wave resonances inside the tunnel. Using your subsonic amplifier and loudspeaker, you find resonances at \(4.5 \mathrm{Hz}\) and 6.3 Hz, and at no frequencies between these. It's rather chilly inside the tunnel, so you estimate the sound speed to be \(335 \mathrm{m} / \mathrm{s}\). Based on your measurements, how far is it to the end of the tunnel?

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