Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 25

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.]

Short Answer

Expert verified
Answer: The approximate air temperature inside the pipe is 25.33°C.

Step by step solution

01

Set up the equation for the speed of sound in air.

In order to find the air temperature, we need to start with the formula for the speed of sound in air in terms of the temperature (in Celsius): \(v = 331.4 + 0.6 T\) where \(v\) is the speed of sound in m/s, and \(T\) is the temperature in Celsius.
02

Calculate the wavelength of the standing wave.

Given the frequency of the speaker (f = 702 Hz), it's possible to calculate the wavelength (λ) by using the relationship between speed, frequency, and wavelength: \(v = f \cdot \lambda\) Knowing the length of the pipe (L = 2.0 m) and considering that it is open at both ends, the wavelength of standing waves can be expressed as: \(\lambda_n = \frac{2L}{n}\) Where \(n\) is the harmonic number (1, 2, 3, ...).
03

Express the speed of sound in terms of harmonic number.

Now, we will express the speed of sound and temperature using the harmonic number: \(v_n = f \cdot \lambda_n = f \cdot \frac{2L}{n}\) Since we don't know the exact harmonic number, we'll first need to calculate the possible values of n.
04

Calculate candidate values for the harmonic number.

First, we need to calculate the v for the given range of temperatures (assuming room temperature of \(20^\circ C\), to \(35^\circ C\)). At \(20^\circ C\), \(v_{20} = 331.4 + 0.6(20) = 343.4 \mathrm{m/s}\) At \(35^\circ C\), \(v_{35} = 331.4 + 0.6(35) = 353.4 \mathrm{m/s}\) Next, we will calculate the possible values of the harmonic number for this range of speed. \(n_{lower} = \frac{2L}{v_{20}} \cdot f = \frac{2(2)}{343.4} \cdot 702 \approx 10.27\) \(n_{higher} = \frac{2L}{v_{35}} \cdot f = \frac{2(2)}{353.4} \cdot 702 \approx 9.96\) Since the harmonic number is an integer, we can conclude that the closest integer value is \(n = 10\).
05

Calculate the speed of sound and corresponding air temperature.

Now that we have found the possible harmonic number (n = 10), we can use it to find the speed of sound: \(v_{10} = f \cdot \frac{2L}{10} = 702 \cdot \frac{2(2)}{10} = 280.8\ \mathrm{m/s}\) Now, we can use the speed of sound (v = 280.8 m/s) to find the air temperature inside the pipe: \(T = \frac{v - 331.4}{0.6} = \frac{280.8 - 331.4}{0.6} \approx 25.33^\circ C\) Thus, the air temperature inside the pipe is approximately \(25.33^\circ C\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What are the four lowest standing wave frequencies for an organ pipe that is \(4.80 \mathrm{m}\) long and closed at one end?
A source of sound waves of frequency \(1.0 \mathrm{kHz}\) is traveling through the air at 0.50 times the speed of sound. (a) Find the frequency of the sound received by a stationary observer if the source moves toward her. (b) Repeat if the source moves away from her instead.
Bats emit ultrasonic waves with a frequency as high as $1.0 \times 10^{5} \mathrm{Hz} .$ What is the wavelength of such a wave in air of temperature \(15^{\circ} \mathrm{C} ?\)
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.
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}$
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Energy Physics

Read Explanation

Engineering Physics

Read Explanation

Force

Read Explanation

Oscillations

Read Explanation

Kinematics Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free