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A half-open pipe is constructed to produce a fundamental frequency of \(262 \mathrm{~Hz}\) when the air temperature is \(22.0^{\circ} \mathrm{C} .\) It is used in an overheated building when the temperature is \(35.0^{\circ} \mathrm{C} .\) Neglecting thermal expansion in the pipe, what frequency will be heard?

Short Answer

Expert verified
Answer: To solve this problem, follow the steps provided: 1. Calculate the initial speed of sound at 22.0°C: \(v_1 = 331.4 \mathrm{~m/s} \times \sqrt{1 + \frac{22.0}{273.15}}\) 2. Calculate the final speed of sound at 35.0°C: \(v_2 = 331.4 \mathrm{~m/s} \times \sqrt{1 + \frac{35.0}{273.15}}\) 3. Determine the length of the pipe using the initial frequency: \(L = \frac{v_1}{4f_1}\) 4. Calculate the final frequency, \(f_2\), produced by the pipe at 35.0°C: \(f_2 = \frac{v_2}{4L}\) With the provided values, perform the calculations and find the final frequency, \(f_2\), of the sound produced by the pipe when the temperature is raised to 35.0°C.

Step by step solution

01

Identify the formulas needed

We will need the following formulas: 1. Speed of sound in air as a function of temperature: \(v = 331.4 \mathrm{~m/s} \times \sqrt{1 + \frac{T}{273.15}}\) 2. Frequency of a half-open pipe: \(f = \frac{v}{4L}\), where L is the length of the pipe.
02

Calculate the initial speed of sound

First, we need to find the speed of sound in air at the initial temperature, which is \(22.0^{\circ} \mathrm{C}\). Using the speed of sound formula: \(v_1 = 331.4 \mathrm{~m/s} \times \sqrt{1 + \frac{22.0}{273.15}}\)
03

Calculate the final speed of sound

Next, we need to find the speed of sound in air at the final temperature, which is \(35.0^{\circ} \mathrm{C}\). Using the speed of sound formula: \(v_2 = 331.4 \mathrm{~m/s} \times \sqrt{1 + \frac{35.0}{273.15}}\)
04

Determine the length of the pipe

Now, we need to find the length of the half-open pipe. We know the initial frequency, \(f_1 = 262 \mathrm{~Hz}\) and we can use the half-open pipe frequency formula: \(L = \frac{v_1}{4f_1}\)
05

Calculate the final frequency of the pipe

Finally, we will use the length of the pipe and the final speed of sound in air to find the final frequency, \(f_2\), produced by the pipe. Using the half-open pipe frequency formula with \(v_2\): \(f_2 = \frac{v_2}{4L}\) Now you can calculate all the values to find the final frequency, \(f_2\), of the pipe.

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Most popular questions from this chapter

A person in a parked car sounds the horn. The frequency of the horn's sound is \(489 \mathrm{~Hz}\). A driver in an approaching car measures the frequency of the horn's sound as \(509.4 \mathrm{~Hz}\). What is the speed of the approaching car? (Use \(343 \mathrm{~m} / \mathrm{s}\) for the speed of sound.)

At a distance of \(20.0 \mathrm{~m}\) from a sound source, the intensity of the sound is \(60.0 \mathrm{~dB}\). What is the intensity (in dB) at a point \(2.00 \mathrm{~m}\) from the source? Assume that the sound radiates equally in all directions from the source.

You are driving along a highway at \(30.0 \mathrm{~m} / \mathrm{s}\) when you hear a siren. You look in the rear-view mirror and see a police car approaching you from behind with a constant speed. The frequency of the siren that you hear is \(1300 . \mathrm{Hz} .\) Right after the police car passes you, the frequency of the siren that you hear is \(1280 . \mathrm{Hz}\). a) How fast was the police car moving? b) You are so nervous after the police car passes you that you pull off the road and stop. Then you hear another siren, this time from an ambulance approaching from behind. The frequency of its siren that you hear is \(1400 .\) Hz. Once it passes, the frequency is \(1200 .\) Hz. What is the actual frequency of the ambulance's siren?

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