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Chapter 18: Problem 50

An aluminum rod is clamped one quarter of the way along its length and set into longitudinal vibration by a variablefrequency driving source. The lowest frequency that produces resonance is \(4400 \mathrm{Hz}\). The speed of sound in an aluminum rod is \(5100 \mathrm{m} / \mathrm{s}\). Find the length of the rod.

Short Answer

Expert verified
Taking resonance condition into account, the length of the rod is found to be approximately \(1.159 m\).

Step by step solution

01

Understand the resonance condition of the rod

The rod is clamped one quarter of the way. This implies that, at the lowest resonant frequency, the rod vibrates in such a way that it forms an antinode at the clamp point and nodes at the ends. In other words, a quarter of the rod's total length represents a wavelength quarter of the resonant wave.
02

Express the wavelength in terms of rod length

According to Step 1, the total length of the rod, let's denote it by \(L\), equals four times the length of a quarter-wavelength, which we can denote by \(\lambda / 4\). We can therefore write the following equation: \(L=4 * \lambda / 4\), and after simplifying this equation, we get \(\lambda = L\).
03

Find the wavelength using the speed of sound and frequency

The equation for wave speed is \(v=f * \lambda\), where \(v\) is the speed of sound, \(f\) is the frequency, and \(\lambda\) is the wavelength. With the known values of \(v = 5100 m/s\) and \(f = 4400 Hz\), we can solve for \(\lambda\): \(\lambda = v/f = 5100 m/s / 4400 Hz = 1.159 m\).
04

Determine the length of the rod

Since the wavelength \(\lambda\) equals the length of the rod \(L\) from Step 2, we also find \(L = 1.159 m\). This is the length of the rod.

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Key Concepts

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

Longitudinal Vibrations
When we talk about longitudinal vibrations, we're referring to oscillations where the displacement of particles in the material occurs in the same direction as the wave propagation. In simpler terms, if you imagine a slinky stretched out along a table, a longitudinal vibration would be created by pushing and pulling the slinky along the length of the table, causing compressions and rarefactions to move along its length.

This type of vibration is exactly what happens in a rod when it undergoes resonance at its fundamental frequency. The particles within the rod move back and forth along the direction of the rod's length, producing sound waves that we can hear. The vibrations also manifest in the form of standing waves with nodes, where there is no movement, and antinodes, where the movement is maximum. When a rod is clamped at a certain point, that point acts as an antinode, and the nodes form at the ends as well as at other points depending on the mode of vibration.
Speed of Sound in Solids
The speed of sound is the rate at which sound waves travel through a medium. The rate is much faster in solids like aluminum than in air because particles in solids are closer together, allowing them to transmit vibrations more quickly. This transmission of vibrational energy governs how quickly sounds, or in this case, resonance frequencies move through a solid material.

For instance, the given exercise states a speed of sound in an aluminum rod as 5100 m/s. This information is crucial since it ties directly into the formula used to find the wavelength, and ultimately the resonant length of the rod. The faster the speed of sound in a material, the greater the distance a sound wave can travel in the same amount of time, affecting the wavelength and resonant frequencies.
Wavelength Calculation
Wavelength, denoted by \(\lambda\), is the distance between successive points of a wave in which the oscillation is at the same phase. In the context of the problem, wavelength calculation is essential for understanding the physical length of a wave within a medium - in our case, the aluminum rod.

By applying the formula \(\lambda = \frac{v}{f}\), where \(v\) represents the speed of sound in the material and \(f\) is the frequency of the longitudinal wave, we can calculate the wavelength. This relationship allows us to determine not just the wavelength but also the length of the rod, since in the fundamental mode of vibration, the wavelength is equal to the length of the rod, assuming one end is free and the other end is an antinode due to clamping. Understanding how to calculate wavelength is an important skill in physics, as it connects the concepts of wave speed and frequency to the spatial configuration of a wave within a material.

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Most popular questions from this chapter

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 \(5.00 \mathrm{Hz} ?\)

A cello A-string vibrates in its first normal mode with a frequency of \(220 \mathrm{Hz}\). The vibrating segment is \(70.0 \mathrm{cm}\) long and has a mass of \(1.20 \mathrm{g} .\) (a) Find the tension in the string. (b) Determine the frequency of vibration when the string vibrates in three segments.

The Bay of Fundy, Nova Scotia, has the highest tides in the world, as suggested in the photographs on page \(452 .\) Assume that in mid-ocean and at the mouth of the bay, the Moon's gravity gradient and the Earth's rotation make the water surface oscillate with an amplitude of a few centimeters and a period of 12 h 24 min. At the head of the bay, the amplitude is several meters. Argue for or against the proposition that the tide is amplified by standing-wave resonance. Assume the bay has a length of \(210 \mathrm{km}\) and a uniform depth of \(36.1 \mathrm{m} .\) The speed of long-wavelength water waves is given by \(\sqrt{g} d\), where \(d\) is the water's depth.

Two speakers are driven in phase by a common oscillator at \(800 \mathrm{Hz}\) and face each other at a distance of \(1.25 \mathrm{m}\) Locate the points along a line joining the two speakers where relative minima of sound pressure amplitude would be expected. (Use \(v=343 \mathrm{m} / \mathrm{s} .\) )

Two pulses traveling on the same string are described by $$y_{1}=\frac{5}{(3 x-4 t)^{2}+2} \quad $$ and $$\quad y_{2}=\frac{-5}{(3 x+4 t-6)^{2}+2}$$ (a) In which direction does each pulse travel? (b) At what time do the two cancel everywhere? (c) At what point do the two pulses always cancel?
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