Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Question: (a) By what factor must the sound intensity be increased to raise the sound intensity level by 13.0 dB? (b) Explain why you don’t need to know the original sound intensity.

Short Answer

Expert verified

(a) the factor for the sound intensity to be increased to raise the sound intensity level by 13.0 dB is 20.

Step by step solution

01

Given data

The increase in sound level intensity is 13 dB.

02

Concept

Sound intensity can be measured as the number of wave particles falling on the surface that can be measured in 1 second.

03

(a) Calculation of the factor

The sound intensity level was \({\beta _1}\) , and it got increased, and the value can be written as,\({\beta _2}\).

The formula for the initial level is,

\({\beta _1} = 10\log \left( {\frac{{{I_1}}}{{{I_0}}}} \right)\)

Here \({I_0}\) is the original intensity, \({I_1}\)is the measured intensity.

The expression for final intensity level is,

\({\beta _2} = 10\log \left( {\frac{{{I_2}}}{{{I_0}}}} \right)\)

Here \({I_2}\)is the measured intensity.

The increment or change is 13 dB; we can write it as,

\(\Delta \beta = {\beta _2} - {\beta _1}\)

Substitute the values in the above expression, and we get,

\(\begin{array}{c}13 = 10\log \left( {\frac{{{I_2}}}{{{I_0}}}} \right) - 10\log \left( {\frac{{{I_1}}}{{{I_0}}}} \right)\\13 = 10\log \left( {\frac{{\frac{{{I_2}}}{{{I_0}}}}}{{\frac{{{I_1}}}{{{I_0}}}}}} \right)\end{array}\)

Solving further as,

\(\begin{array}{c}13 = 10\log \left( {\frac{{{I_2}}}{{{I_1}}}} \right)\\1.3 = \log \left( {\frac{{{I_2}}}{{{I_1}}}} \right)\\\frac{{{I_2}}}{{{I_1}}} = 20\end{array}\)

Thus, the factor for the sound intensity to be increased to raise the sound intensity level by 13.0 dB is 20.

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.

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

As we discussed in Section 15.1, water waves are a combination of longitudinal and transverse waves. Defend the following statement: “When water waves hit a vertical wall, the wall is a node of the longitudinal displacement but an antinode of the transverse displacement.”

Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750 Hz. (a) What frequency note does each bird hear from the other one? (b) What wavelength will each canary measure for the note from the other one?

15.4. BIO Ultrasound Imaging. Sound having frequencies above the range of human hearing (about 20,000 Hz) is called ultrasound. Waves above this frequency can be used to penetrate the body and to produce images by reflecting from surfaces. In a typical ultrasound scan, the waves travel through body tissue with a speed of 1500 m/s . For a good, detailed image, the wavelength should be no more than 1.0 mm. What frequency sound is required for a good scan?

A long tube contains air at a pressure of 1.00 atm and a temperature of 77.0°C. The tube is open at one end and closed at the other by a movable piston. A tuning fork that vibrates with a frequency of 500 Hz is placed near the open end. Resonance is produced when the piston is at distances 18.0 cm, 55.5 cm, and 93.0 cm from the open end. (a) From these values, what is the speed of sound in air at 77.0°C? (b) From the result of part (a), what is the value of g? (c) These results show that a displacement antinode is slightly outside the open end of the tube. How far outside is it?

In most modern wind instruments the pitch is changed by

using keys or valves to change the length of the vibrating air column.

The bugle, however, has no valves or keys, yet it can play many notes. How might this be possible? Are there restrictions on what notes a bugle can play?

See all solutions

Recommended explanations on Physics Textbooks

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