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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

In an experiment to measure the speed of sound in air, standing waves are set up in a narrow pipe open at both ends using a speaker driven at \(702 Hz\). The length of the pipe is \(2.0 \mathrm{m} .\) What is the air temperature inside the pipe (assumed reasonably near room temperature, \(20^{\circ} \mathrm{C}\) to \(\left.35^{\circ} \mathrm{C}\right) ?\) [Hint: The standing wave is not necessarily the fundamental.]
A geological survey ship mapping the floor of the ocean sends sound pulses down from the surface and measures the time taken for the echo to return. How deep is the ocean at a point where the echo time (down and back) is $7.07 \mathrm{s} ?\( The temperature of the seawater is \)25^{\circ} \mathrm{C}$
Humans can hear sounds with frequencies up to about \(20.0 \mathrm{kHz},\) but dogs can hear frequencies up to about \(40.0 \mathrm{kHz} .\) Dog whistles are made to emit sounds that dogs can hear but humans cannot. If the part of a dog whistle that actually produces the high frequency is made of a tube open at both ends, what is the longest possible length for the tube?
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?
Some bats determine their distance to an object by detecting the difference in intensity between cchoes.(a) If intensity falls off at a rate that is inversely proportional to the distance squared, show that the echo intensity is inversely proportional to the fourth power of distance. (b) The bat was originally \(0.60 \mathrm{m}\) from one object and \(1.10 \mathrm{m}\) from another. After flying closer, it is now \(0.50 \mathrm{m}\) from the first and at \(1.00 \mathrm{m}\) from the second object. What is the percentage increase in the intensity of the ccho from each object?
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