An aluminum rod, \(1.0 \mathrm{m}\) long, is held lightly in the middle. One end is struck head-on with a rubber mallet so that a longitudinal pulse-a sound wave- -travels down the rod. The fundamental frequency of the longitudinal vibration is \(2.55 \mathrm{kHz}\). (a) Describe the location of the node(s) and antinode(s) for the fundamental mode of vibration. Use either displacement or pressure nodes and antinodes. (b) Calculate the speed of sound in aluminum from the information given in the problem. (c) The vibration of the rod produces a sound wave in air that can be heard. What is the wavelength of the sound wave in the air? Take the speed of sound in air to be \(334 \mathrm{m} / \mathrm{s}\). (d) Do the two ends of the rod vibrate longitudinally in phase or out of phase with each other? That is, at any given instant, do they move in the same direction or in opposite directions?

Step by step solution

01

Identifying nodes and antinodes for the fundamental mode of vibration

In the fundamental mode of vibration, there is one node at the center of the rod and two antinodes at the ends of the rod. The displacement antinode is the point of maximum displacement, and the displacement node is the point of no displacement.

02

Calculating the speed of sound in aluminum

With the fundamental frequency \(f = 2.55 \mathrm{kHz}\), and the aluminum rod being 0.5m long (half of the full length since it's held in the middle), we can determine the speed of sound in aluminum. The speed of sound in a medium is given by the formula: \(v = f\lambda\) To find the speed of sound, we need to determine the wavelength first. For the fundamental mode of vibration, the wavelength (λ) is twice the length of the rod: \(\lambda = 2L = 2(0.5\,\mathrm{m}) = 1\,\mathrm{m}\) Now, we can calculate the speed of sound in aluminum: \(v = (2.55 \,\mathrm{kHz})(1\,\mathrm{m}) = 2550\, \mathrm{m}/\mathrm{s}\)

03

Calculating the wavelength of sound wave in air

The speed of sound in air is given as \(334\, \mathrm{m}/\mathrm{s}\). We can find the wavelength of the sound wave in air by using the same speed of sound formula: \(v = f\lambda\) We can solve for the wavelength in air: \(\lambda_{air} = \frac{v}{f} = \frac{334\, \mathrm{m}/\mathrm{s}}{2.55\, \mathrm{kHz}} = \frac{334}{2550}\, \mathrm{m} \approx 0.131\,\mathrm{m}\)

04

Determining the phase of the ends of the rod

In the fundamental mode of vibration, the ends of the rod (antinodes) are vibrating in opposite directions (longitudinally). This means that they are out of phase with each other. When one end is at maximum displacement in one direction, the other end is at maximum displacement in the opposite direction. In summary: a) There is one node at the center of the rod and two antinodes at the ends of the rod. b) The speed of sound in aluminum is \(2550\, \mathrm{m}/\mathrm{s}\). c) The wavelength of the sound wave in air is approximately \(0.131\, \mathrm{m}\). d) The ends of the rod vibrate out of phase with each other.

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