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Chapter 12: Problem 22

An organ pipe that is open at both ends has a fundamental frequency of $382 \mathrm{Hz}\( at \)0.0^{\circ} \mathrm{C}$. What is the fundamental frequency for this pipe at \(20.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The fundamental frequency of the open-ended organ pipe at 20 degrees Celsius is approximately 396 Hz.

Step by step solution

01

Find the speed of sound at 0 degrees Celsius

First, we need to find the speed of sound in the air at 0 degrees Celsius. We can use the following formula to find the speed of sound in the air: \(v = 331.4 \sqrt{1+\frac{T}{273.15}}\) Where: \(v\) is the speed of sound in meters per second (m/s) \(T\) is the temperature in Celsius For this problem, \(T = 0.0^{\circ} \mathrm{C}\), so we plug this into the equation: \(v = 331.4 \sqrt{1+\frac{0}{273.15}} \approx 331.4 \, \mathrm{m/s}\)
02

Calculate the wavelength of the fundamental frequency at 0 degrees Celsius

Next, we will find the wavelength of the fundamental frequency at 0 degrees Celsius. To find the wavelength of the fundamental frequency, we will use the following equation: \(\lambda = \frac{v}{f}\) Where: \(\lambda\) is the wavelength in meters (m) \(v\) is the speed of sound in meters per second (m/s) \(f\) is the frequency in Hertz (Hz) For this problem, the frequency at 0 degrees Celsius is 382 Hz. So, we plug this into the equation: \(\lambda = \frac{331.4 \, \mathrm{m/s}}{382 \, \mathrm{Hz}} \approx 0.867 \, \mathrm{m}\)
03

Find the speed of sound at 20 degrees Celsius

Now, we need to find the speed of sound in the air at 20 degrees Celsius. We use the same formula as in step 1 with \(T = 20.0^{\circ} \mathrm{C}\): \(v = 331.4 \sqrt{1+\frac{20}{273.15}} \approx 343.4 \, \mathrm{m/s}\)
04

Calculate the fundamental frequency at 20 degrees Celsius

Finally, we can now find the fundamental frequency at 20 degrees Celsius using the new speed of sound and the same wavelength as calculated in step 2: \(f = \frac{v}{\lambda}\) Plugging in the values we have: \(f = \frac{343.4 \, \mathrm{m/s}}{0.867 \, \mathrm{m}} \approx 396 \, \mathrm{Hz}\) So, the fundamental frequency for the organ pipe at \(20.0^{\circ} \mathrm{C}\) is approximately \(396 \, \mathrm{Hz}\).

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Most popular questions from this chapter

A piano tuner sounds two strings simultancously. One has been previously tuned to vibrate at \(293.0 \mathrm{Hz}\). The tuner hears 3.0 beats per second. The tuner increases the tension on the as-yet untuned string, and now when they are played together the beat frequency is \(1.0 \mathrm{s}^{-1} .\) (a) What was the original frequency of the untuned string? (b) By what percentage did the tuner increase the tension on that string?
In this problem, you will estimate the smallest kinetic energy of vibration that the human ear can detect. Suppose that a harmonic sound wave at the threshold of hearing $\left(I=1.0 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}\right)\( is incident on the eardrum. The speed of sound is \)340 \mathrm{m} / \mathrm{s}\( and the density of air is \)1.3 \mathrm{kg} / \mathrm{m}^{3} .$ (a) What is the maximum speed of an element of air in the sound wave? [Hint: See Eq. \((10-21) .]\) (b) Assume the eardrum vibrates with displacement \(s_{0}\) at angular frequency \(\omega ;\) its maximum speed is then equal to the maximum speed of an air element. The mass of the eardrum is approximately \(0.1 \mathrm{g} .\) What is the average kinetic energy of the eardrum? (c) The average kinetic energy of the eardrum due to collisions with air molecules in the absence of a sound wave is about \(10^{-20} \mathrm{J}\) Compare your answer with (b) and discuss.
According to a treasure map, a treasure lies at a depth of 40.0 fathoms on the ocean floor due east from the lighthouse. The treasure hunters use sonar to find where the depth is 40.0 fathoms as they head cast from the lighthouse. What is the elapsed time between an emitted pulse and the return of its echo at the correct depth if the water temperature is $\left.25^{\circ} \mathrm{C} ? \text { [Hint: One fathom is } 1.83 \mathrm{m} .\right]$
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An auditorium has organ pipes at the front and at the rear of the hall. Two identical pipes, one at the front and one at the back, have fundamental frequencies of \(264.0 \mathrm{Hz}\) at \(20.0^{\circ} \mathrm{C} .\) During a performance, the organ pipes at the back of the hall are at $25.0^{\circ} \mathrm{C},\( while those at the front are still at \)20.0^{\circ} \mathrm{C} .$ What is the beat frequency when the two pipes sound simultaneously?
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