Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 18, Problem 46P
To determine

The frequencies, which would sound the richest because of resonance.

Expert Solution & Answer
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Explanation of Solution

Given info: The dimension of the shower stall is 86.0cm×86.0cm×210cm . The range of frequencies of the voices lies from 130Hz to 2000Hz . The speed of the sound in the hot air is 355m/s .

Expression for the fundamental or first harmonic frequency is,

f1=v4L

Here,

v is the velocity of the sound in hot air.

L is the length of the shower stall.

Substitute 355m/s for v and 210cm for L in the above equation.

f1=355m/s4×(210cm)=355m/s4×(210cm×1m100cm)=355m/s4×(2.10m)=42.262Hz

Expression for the third harmonic frequency is,

f3=3v4L

Substitute 355m/s for v and 210cm for L in the above equation.

f3=3×(355m/s)4×(210cm)=3×(355m/s)4×(210cm×1m100cm)=3×(355m/s)4×(2.10m)=126.786Hz

Expression for the fifth harmonic frequency is,

f5=5v4L

Substitute 355m/s for v and 210cm for L in the above equation.

f5=5×(355m/s)4×(210cm)=5×(355m/s)4×(210cm×1m100cm)=5×(355m/s)4×(2.10m)=211.31Hz

Expression for the seventh harmonic frequency is,

f7=7v4L

Substitute 355m/s for v and 210cm for L in the above equation.

f7=7×(355m/s)4×(210cm)=7×(355m/s)4×(210cm×1m100cm)=7×(355m/s)4×(2.10m)=295.834Hz

Expression for the ninth harmonic frequency is,

f9=9v4L

Substitute 355m/s for v and 210cm for L in the above equation.

f9=9×(355m/s)4×(210cm)=9×(355m/s)4×(210cm×1m100cm)=9×(355m/s)4×(2.10m)=380.358Hz

Expression for the eleventh harmonic frequency is,

f11=11v4L

Substitute 355m/s for v and 210cm for L in the above equation.

f11=11×(355m/s)4×(210cm)=11×(355m/s)4×(210cm×1m100cm)=11×(355m/s)4×(2.10m)=464.881Hz

Similarly, for maximum resonance frequency,

fn=(2n+1)v4L

Substitute 2000Hz for fn , 355m/s for v and 210cm for L in the above equation.

2000Hz=(2n+1)×(355m/s)4×(210cm)2000Hz=(2n+1)×(355m/s)4×(210cm×1m100cm)=(2n+1)×(355m/s)4×(210cm×1m100cm)

Simplify further,

2000Hz=(2n+1)×(355m/s)4×(2.10m)n47

For n=47 ,

f47=47×(42.262Hz)=1986.314Hz

Conclusion:

Therefore, the frequencies, which would sound the richest because of resonance, are 211.31Hz , 295.834Hz , 380.358Hz , 464.881Hz up to 1986.314Hz .

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Chapter 18 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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