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Chapter 17: Problem 50

A tube \(1.20 \mathrm{~m}\) long is closed at one end. A stretched wire is placed near the open end. The wire is \(0.330 \mathrm{~m}\) long and has a mass of \(9.60 \mathrm{~g} .\) It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Find (a) that frequency and (b) the tension in the wire.

Short Answer

Expert verified
(a) 70.83 Hz, (b) 63.63 N

Step by step solution

01

Understand Resonance in a Closed Tube

For a tube closed at one end, the fundamental frequency occurs when there is a quarter wavelength (\(\frac{\lambda}{4}\)) inside the tube. The length of the tube (\(L\)) is given as 1.2 m, so the relationship is \[ L = \frac{\lambda}{4} \] Therefore, the wavelength \(\lambda\) is 4 times the length of the tube: \[ \lambda = 4 \times L = 4 \times 1.2 = 4.8 \text{ m} \]
02

Calculate Fundamental Frequency in Closed Tube

The fundamental frequency \(f\) of a wave is given by: \[ f = \frac{v}{\lambda} \] where \(v\) is the speed of sound in air, typically around 340 m/s (assuming normal conditions). Using the wavelength from Step 1: \[ f = \frac{340 \text{ m/s}}{4.8 \text{ m}} \approx 70.83 \text{ Hz} \]
03

Find Frequency of Wire at Fundamental Mode

The frequency of the wire oscillating in its fundamental mode is the same as the fundamental frequency of the air column due to resonance: \(f = 70.83 \text{ Hz}\)
04

Understand Wire Resonance and Calculate Speed

For the wire oscillating in its fundamental mode, the length of the wire \(L = 0.330 \text{ m}\) is half a wavelength: \(L = \frac{\lambda_{wire}}{2}\). Therefore, \(\lambda_{wire} = 2 \times 0.330 = 0.660 \text{ m}\). The wave speed \(v_{wire}\) is related to frequency by: \[ v_{wire} = f \times \lambda_{wire} \] Using \(f = 70.83 \text{ Hz}\): \[ v_{wire} = 70.83 \text{ Hz} \times 0.660 \text{ m} \approx 46.75 \text{ m/s} \]
05

Calculate Tension in the Wire

The speed of waves on a string is given by: \[ v = \sqrt{\frac{T}{\mu}} \] where \(T\) is the tension, and \(\mu = \frac{\text{mass}}{\text{length}} = \frac{9.60 \text{ g}}{0.330 \text { m}} = 0.0291 \text{ kg/m}\) (because 9.60 grams is 0.0096 kg). Solving for \(T\): \[ T = v_{wire}^2 \times \mu = (46.75 \text{ m/s})^2 \times 0.0291 \text{ kg/m} \approx 63.63 \text{ N} \]

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Key Concepts

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

Fundamental frequency
Resonance occurs in a closed tube when the air column inside vibrates at its natural frequencies, creating standing waves. The fundamental frequency is the lowest frequency at which resonance occurs. In a tube closed at one end, this fundamental frequency resonates when the length of the tube accommodates one quarter of the wave's wavelength.
  • The length of the tube here is 1.20 m.
  • This means that the length corresponds to \(\frac{\lambda}{4}=1.20\text{ m}\).
  • Rearranging gives the wavelength \(\lambda=4\times 1.20=4.8\text{ m}\).
Furthermore, the fundamental frequency \(f\) can be calculated with the wave speed \(v\) and wavelength \(\lambda\) using the equation: \(f=\frac{v}{\lambda}\). With the speed of sound in air given as 340 m/s, the fundamental frequency of the tube is approximately 70.83 Hz, which matches the resonant frequency of the wire.
Wave speed
Wave speed refers to how quickly a wave travels through a medium, determined by its frequency and wavelength. When a wire oscillates at its fundamental frequency, the wave speed is the same as the product of its frequency and its wavelength on the wire.For our wire:
  • Its length is 0.330 m, representing half the wavelength of the standing wave, and hence, the total wavelength \(\lambda_{wire}=2\times 0.330=0.660 \text{ m}\).
  • Knowing its frequency is 70.83 Hz (as it matches the tube's resonant frequency), we compute the wave speed on the wire using \(v_{wire}=f\times \lambda_{wire}=70.83 \text{ Hz}\times 0.660 \text{ m}\approx 46.75 \text{ m/s}\).
This speed signifies how fast wave crests are moving along the wire as it vibrates.
Tension in the wire
The tension in a string or wire affects how fast waves can travel along it. This speed can be computed from the tension and linear mass density of the wire. The formula to find the wave speed, based on tension \(T\) and linear mass density \(\mu\), is \(v=\sqrt{\frac{T}{\mu}}\).The linear mass density \(\mu\) is calculated by dividing the mass of the wire by its length:
  • Mass = 9.60 g = 0.0096 kg
  • Length = 0.330 m
  • \(\mu=\frac{0.0096 \text{ kg}}{0.330 \text{ m}}=0.0291 \text{ kg/m}\)
To find the tension, rearrange the speed formula: \(T=v_{wire}^2\times \mu\). Substituting the wave speed \(v_{wire}\approx 46.75 \text{ m/s}\), we find \(T\approx (46.75 \text{ m/s})^2\times 0.0291 \text{ kg/m}\approx 63.63 \text{ N}\). This tension allows the wire to vibrate at its resonant frequency.
Wavelength in tubes
Wavelength is the distance over which the wave's shape repeats. In closed tubes, the wavelength determines the resonating frequencies. For a tube closed at one end:The tube allows specific wavelengths that have a node at the closed end and an antinode at the open end. This setup means a quarter of a wavelength fits into the length of the tube for its fundamental frequency.
  • With a tube length of 1.20 m, as seen in this case, this is \(\frac{\lambda}{4}=1.20 \text{ m}\).
  • Thus, the entire wavelength is \(\lambda=4\times 1.20=4.8 \text{ m}\).
This calculation gives the lowest resonant wavelength supporting a standing wave with these specific boundary conditions, crucial for the fundamental frequency.

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Most popular questions from this chapter

A whistle of frequency \(540 \mathrm{~Hz}\) moves in a circle of radius \(60.0 \mathrm{~cm}\) at an angular speed of \(15.0 \mathrm{rad} / \mathrm{s}\). What are the (a) lowest and (b) highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle?

A sound wave travels out uniformly in all directions from a point source. (a) Justify the following expression for the displacement \(s\) of the transmitting medium at any distance \(r\) from the source: $$ s=\frac{b}{r} \sin k(r-v t) $$ where \(b\) is a constant. Consider the speed, direction of propagation, periodicity, and intensity of the wave. (b) What is the dimension of the constant \(b ?\)

Four sound waves are to be sent through the same tube of air, in the same direction: $$ \begin{array}{l} s_{1}(x, t)=(9.00 \mathrm{~nm}) \cos (2 \pi x-700 \pi t) \\ s_{2}(x, t)=(9.00 \mathrm{~nm}) \cos (2 \pi x-700 \pi t+0.7 \pi) \\ s_{3}(x, t)=(9.00 \mathrm{~nm}) \cos (2 \pi x-700 \pi t+\pi) \\ s_{4}(x, t)=(9.00 \mathrm{~nm}) \cos (2 \pi x-700 \pi t+1.7 \pi) . \end{array} $$ What is the amplitude of the resultant wave? (Hint: Use a phasor diagram to simplify the problem.)

Tap a metal spoon inside a mug of water and note the frequency \(f_{i}\) you hear. Then add a spoonful of powder (say, chocolate mix or instant coffee) and tap again as you stir the powder. The frequency you hear has a lower value \(f_{s}\) because the tiny air bubbles released by the powder change the water's bulk modulus. As the bubbles reach the water surface and disappear, the frequency gradually shifts back to its initial value. During the effect, the bubbles don't appreciably change the water's density or volume or the sound's wavelength. Rather, they change the value of \(d V / d p-\) that is, the differential change in volume due to the differential change in the pressure caused by the sound wave in the water. If \(f_{s} / f_{i}=0.333\), what is the ratio \((d V / d p)_{s} /(d V / d p)_{i} ?\)

The source of a sound wave has a power of \(1.00 \mu \mathrm{W}\). If it is a point source, (a) what is the intensity \(3,00 \mathrm{~m}\) away and (b) what is the sound level in decibels at that distance?
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