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Chapter 16: Problem 50

A thin aluminum rod of length \(L=2.00 \mathrm{~m}\) is clamped at its center. The speed of sound in aluminum is \(5000 . \mathrm{m} / \mathrm{s}\). Find the lowest resonance frequency for vibrations in this rod.

Short Answer

Expert verified
Answer: The lowest resonance frequency for the given aluminum rod is 1250 Hz.

Step by step solution

01

Identify given values

We are given the length of the aluminum rod, \(L = 2.00 \mathrm{~m}\), and the speed of sound in aluminum, \(v = 5000 \mathrm{~m/s}\).
02

Apply the formula for the fundamental frequency

We will use the formula for the fundamental frequency of a freely vibrating rod clamped at its center: \(f_1 = \frac{v}{2L}\). Substitute the given values of \(v = 5000 \mathrm{~m/s}\) and \(L = 2.00 \mathrm{~m}\) into the formula: \(f_1 = \frac{5000 \mathrm{~m/s}}{2(2.00 \mathrm{~m})}\)
03

Calculate the fundamental frequency

Perform the calculation in the formula: \(f_1 = \frac{5000 \mathrm{~m/s}}{4.00 \mathrm{~m}} = 1250 \mathrm{~Hz}\) Therefore, the lowest resonance frequency for vibrations in the rod is \(1250 \mathrm{~Hz}\).

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Most popular questions from this chapter

A standing wave in a pipe with both ends open has a frequency of \(440 \mathrm{~Hz}\). The next higher harmonic has a frequency of \(660 \mathrm{~Hz}\) a) Determine the fundamental frequency. b) How long is the pipe?

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