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Standing-wave vibrations are set up in a crystal goblet with four nodes and four antinodes equally spaced around the 20.0 -cm circumference of its rim. If transverse waves move around the glass at \(900 \mathrm{m} / \mathrm{s},\) an opera singer would have to produce a high harmonic with what frequency to shatter the glass with a resonant vibration?

Short Answer

Expert verified
The opera singer would have to produce a frequency of \(9000 \mathrm{Hz}\) to shatter the glass with a resonant vibration.

Step by step solution

01

Determine the Wavelength

The number of nodes is four, which is half the number of wavelengths around the rim of the goblet. Therefore, the wavelength of the wave equals twice the circumference divided by the total number of wavelengths, or \(\lambda = 2 \times (20.0 \mathrm{cm} / 4) = 10.0 \mathrm{cm}\). It is important to change this to meters, yielding \(\lambda = 0.10 \mathrm{m}\).
02

Calculate the Frequency

Substituting \(\lambda = 0.10 \mathrm{m}\) and \(v = 900 \mathrm{m/s}\) (the speed of the wave) into the wave velocity equation (v = \(\lambda \times f\)), we can rearrange to find the frequency of the wave: \(f = v / \lambda = 900 \mathrm{m/s} / 0.10 \mathrm{m} = 9000 \mathrm{Hz}\).

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Key Concepts

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

Resonant Vibration
The term 'resonant vibration' is crucial when discussing how a sound wave can cause an object to vibrate at its natural frequency. This concept is important in the context of the opera singer attempting to shatter a glass. When an object naturally vibrates, it does so at its resonant frequency.
If an external force, like a sound wave, matches this frequency, the object will absorb energy more efficiently and vibrate with a greater amplitude. This is known as resonant vibration.
  • In the crystal goblet example, the resonant vibration leads to an increased vibration amplitude.
  • If the amplitude becomes large enough, the material of the glass reaches its breaking point and shatters.
Understanding resonant vibrations helps us appreciate how energy transfers from one source, such as an opera singer's voice, can vividly impact another object, enabling even seemingly implausible feats like glass shattering.
Harmonic Frequency
Harmonic frequencies relate to multiples of a fundamental frequency, creating a harmonic series. In the exercise, the glass is set to vibrate at a specific harmonic frequency by the singer.
Each harmonic frequency corresponds to a standing wave pattern; they are characterized by nodes and antinodes. A node is a point of no displacement, while an antinode is where the maximum displacement occurs.
  • For the goblet, having four nodes and antinodes indicates it's vibrating at a higher harmonic.
  • This means the singer needs to hit a frequency that matches this particular harmonic setup for resonance.
Hitting the right harmonic frequency is critical because only specific frequencies will cause the glass to resonate and potentially vibrate with enough force to fracture.
Opera Singer
An opera singer uses powerful vocal techniques to produce strong, resonant notes. In our example, the singer applies these skills to generate a sound wave that matches the natural harmonic frequency of a glass.
Opera singers can produce a wide range of frequencies and with sufficient training, they can control their voice to hit precise pitches needed to achieve certain effects.
  • The singer must sustain a note at the harmonic frequency corresponding to the resonance of the glass.
  • When successful, the singer's voice provides enough energy absorption by the glass to potentially cause it to shatter.
This showcases how the human voice, when properly trained, can generate forceful impacts on solid objects and brings attention to the power of acoustics in real-world applications.

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Most popular questions from this chapter

The wave function for a standing wave is given in Equation 18.3 as \(y=2 A \sin k x \cos \omega t .\) (a) Rewrite this wave function in terms of the wavelength \(\lambda\) and the wave speed \(v\) of the wave. (b) Write the wave function of the simplest standing wave vibration of a stretched string of length \(L\). (c) Write the wave function for the second harmonic. (d) Generalize these results and write the wave function for the \(n\)th resonance vibration.

Two traveling sinusoidal waves are described by the wave functions $$ y_{1}=(5.00 \mathrm{m}) \sin [\pi(4.00 x-1200 t)] $$ and $$ y_{2}=(5.00 \mathrm{m}) \sin [\pi(4.00 x-1200 t-0.250)] $$ where \(x, y_{1},\) and \(y_{2}\) are in meters and \(t\) is in seconds. (a) What is the amplitude of the resultant wave? (b) What is the frequency of the resultant wave?

An air column in a glass tube is open at one end and closed at the other by a movable piston. The air in the tube is warmed above room temperature, and a \(384-\mathrm{Hz}\) tuning fork is held at the open end. Resonance is heard when the piston is \(22.8 \mathrm{cm}\) from the open end and again when it is \(68.3 \mathrm{cm}\) from the open end. (a) What speed of sound is implied by these data? (b) How far from the open end will the piston be when the next resonance is heard?

With a particular fingering, a flute plays a note with frequency \(880 \mathrm{Hz}\) at \(20.0^{\circ} \mathrm{C} .\) The flute is open at both ends. (a) Find the air column length. (b) Find the frequency it produces at the beginning of the half-time performance at a late-season American football game, when the ambient temperature is \(-5.00^{\circ} \mathrm{C}\) and the musician has not had a chance to warm up the flute.

A shower stall measures \(86.0 \mathrm{cm} \times 86.0 \mathrm{cm} \times 210 \mathrm{cm} .\) If you were singing in this shower, which frequencies would sound the richest (because of resonance)? Assume that the stall acts as a pipe closed at both ends, with nodes at opposite sides. Assume that the voices of various singers range from \(130 \mathrm{Hz}\) to \(2000 \mathrm{Hz}\). Let the speed of sound in the hot shower stall be \(355 \mathrm{m} / \mathrm{s}.\)

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