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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 bat emits chirping sounds of frequency \(82.0 \mathrm{kHz}\) while hunting for moths to eat. If the bat is flying toward the moth at a speed of $4.40 \mathrm{m} / \mathrm{s}\( and the moth is flying away from the bat at \)1.20 \mathrm{m} / \mathrm{s},$ what is the frequency of the sound wave reflected from the moth as observed by the bat? Assume it is a cool night with a temperature of \(10.0^{\circ} \mathrm{C} .\) [Hint: There are two Doppler shifts. Think of the moth as a receiver, which then becomes a source as it "retransmits" the reflected wave. \(]\)
At a baseball game, a spectator is \(60.0 \mathrm{m}\) away from the batter. How long does it take the sound of the bat connecting with the ball to travel to the spectator's ears? The air temperature is \(27.0^{\circ} \mathrm{C}\)
What are the four lowest standing wave frequencies for an organ pipe that is \(4.80 \mathrm{m}\) long and closed at one end?
The Vespertilionidae family of bats detect the distance to an object by timing how long it takes for an emitted signal to reflect off the object and return. Typically they emit sound pulses 3 ms long and 70 ms apart while cruising. (a) If an echo is heard 60 ms later $\left(v_{\text {sound }}=331 \mathrm{m} / \mathrm{s}\right),\( how far away is the object? (b) When an object is only \)30 \mathrm{cm}$ away, how long will it be before the echo is heard? (c) Will the bat be able to detect this echo?
An intensity level change of \(+1.00 \mathrm{dB}\) corresponds to what percentage change in intensity?
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