Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 11, Problem 74P

(a)

To determine

The three lowest standing wave frequencies.

(a)

Expert Solution
Check Mark

Answer to Problem 74P

The three lowest standing wave frequencies is f2=600.0Hz, f3=900.0Hz, and f4=1.200kHz.

Explanation of Solution

Write the equation for lowest standing wave frequency.

fn=nv2L (I)

Here, fn is the fundamental frequency, n is the harmonic number, v is the speed of the sound wave, and L is the length of the string.

Conclusion:

The first lowest fundamental frequency of the standing wave is.

f1=300.0Hz

The second lowest fundamental frequency of the standing wave is.

f2=2f1 (II)

Substitute 300.0Hz for f1 in equation (II).

f2=2(300.0Hz)=600.0Hz

The third lowest fundamental frequency of the standing wave is.

f3=3f1 (III)

Substitute 300.0Hz for f1 in equation (III).

f3=3(300.0Hz)=900.0Hz

The fourth lowest fundamental frequency of the standing wave is.

f4=4f1 (IV)

Substitute 300.0Hz for f1 in equation (IV).

f4=4(300.0Hz)=1200.0Hz=1.200×103Hz(103Hz1kHz)=1.200kHz

The pattern for the each situation for the three lowest standing wave frequencies is.

npatternnf1
2N   A   N   A   N600.0Hz
3N A N A N A N900.0Hz
4NANANANAN1.200Hz

Therefore, the three lowest standing wave frequencies is f2=600.0Hz, f3=900.0Hz, and f4=1.200kHz.

(b)

To determine

The four lowest standing wave frequencies, when the strings press light.

(b)

Expert Solution
Check Mark

Answer to Problem 74P

The four lowest standing wave frequencies is f2=600.0Hz, f4=1.200kHz, f6=1.800kHz, and f8=2.400kHz.

Explanation of Solution

Write the equation for lowest standing wave for fundamental frequency.

fn=nf1 (V)

Here, fn is the fundamental frequency for nth harmonic, n is the harmonic number, and f1 is the fundamental frequency.

Conclusion:

The lowest fundamental frequency of the standing wave is.

f1=300.0Hz

Here, the lowest frequency is starts from f2 and only even harmonics are allowed.

The first lowest fundamental frequency of the standing wave is.

f2=2f1 (VI)

Substitute 300.0Hz for f1 in equation (VI).

f2=2(300.0Hz)=600.0Hz

The second lowest fundamental frequency of the standing wave is.

f4=4f1 (VII)

Substitute 300.0Hz for f1 in equation (VII).

f4=4(300.0Hz)=1200.0Hz=1.200×103Hz(103Hz1kHz)=1.200kHz

The third lowest fundamental frequency of the standing wave is.

f6=6f1 (VIII)

Substitute 300.0Hz for f1 in equation (VIII).

f6=6(300.0Hz)=1800.0Hz=1.800×103Hz(103Hz1kHz)=1.800kHz

The fourth lowest fundamental frequency of the standing wave is.

f8=8f1 (IX)

Substitute 300.0Hz for f1 in equation (IX).

f8=8(300.0Hz)=2400.0Hz=2.400×103Hz(103Hz1kHz)=2.400kHz

The pattern for the each situation for the four lowest standing wave frequencies is.

npatternnf1
2N    A    N    A    N600.0Hz
4N   A   N   A   N   A   N   A   N1.200kHz
6N  A  N  A  N  A  N  A  N  A  N  A  N1.800kHz
8NANANANANANANANAN2.400kHz

Therefore, the four lowest standing wave frequencies is f2=600.0Hz, f4=1.200kHz, f6=1.800kHz, and f8=2.400kHz.

(c)

To determine

The four lowest standing wave frequencies, when the strings press hard.

(c)

Expert Solution
Check Mark

Answer to Problem 74P

The four lowest standing wave frequencies is f1=600.0Hz, f2=1.200kHz, f3=1.800kHz, and f4=2.400kHz.

Explanation of Solution

The effective length of the string is now half of the original length.

Write the equation for lowest standing wave for fundamental frequency.

fn=nv2(L2) (X)

Here, fn is the new fundamental frequency for nth harmonic.

Rearrange the equation (X).

fn=nvL

Substitute equation (I) in above equation.

fn=2fn (XI)

Conclusion:

The first lowest fundamental frequency of the standing wave is.

f1=2(300.0Hz)=600.0Hz

The second lowest fundamental frequency of the standing wave is.

f2=2f2

Substitute 600.0Hz for f2.

f2=2(600.0Hz)=1.200kHz

The third lowest fundamental frequency of the standing wave is.

f3=2f3

Substitute 900.0Hz for f3.

f3=2(900.0Hz)=1800.0Hz=1.800×103Hz(103Hz1kHz)=1.800kHz

The fourth lowest fundamental frequency of the standing wave is.

f4=2f4

Substitute 1.200kHz for f4.

f4=2(1.200kHz)=2.400kHz

The pattern for the each situation for the four lowest standing wave frequencies is.

npatternfn=2fn
1N     A     N600.0Hz
2N    A    N    A    N1.200kHz
3N  A  N  A  N  A  N1.800kHz
4NANANANAN2.400kHz

Therefore, the four lowest standing wave frequencies is f1=600.0Hz, f2=1.200kHz, f3=1.800kHz, and f4=2.400kHz.

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Chapter 11 Solutions

Physics

Ch. 11.9 - Prob. 11.9CPCh. 11.10 - Prob. 11.10CPCh. 11.10 - Prob. 11.7PPCh. 11 - Prob. 1CQCh. 11 - Prob. 2CQCh. 11 - Prob. 3CQCh. 11 - Prob. 4CQCh. 11 - Prob. 5CQCh. 11 - Prob. 6CQCh. 11 - Prob. 7CQCh. 11 - Prob. 8CQCh. 11 - Prob. 9CQCh. 11 - Prob. 10CQCh. 11 - Prob. 11CQCh. 11 - Prob. 1MCQCh. 11 - Prob. 2MCQCh. 11 - Prob. 3MCQCh. 11 - Prob. 4MCQCh. 11 - Prob. 5MCQCh. 11 - Prob. 6MCQCh. 11 - Prob. 7MCQCh. 11 - Prob. 8MCQCh. 11 - Prob. 9MCQCh. 11 - Prob. 10MCQCh. 11 - Prob. 11MCQCh. 11 - Prob. 1PCh. 11 - Prob. 2PCh. 11 - Prob. 3PCh. 11 - Prob. 4PCh. 11 - 5. At what rate does the jet airplane in Problem 4...Ch. 11 - Prob. 6PCh. 11 - Prob. 7PCh. 11 - Prob. 8PCh. 11 - Prob. 9PCh. 11 - Prob. 10PCh. 11 - 11. A metal guitar string has a linear mass...Ch. 11 - Prob. 12PCh. 11 - 13. Two strings, each 15.0 m long, are stretched...Ch. 11 - Prob. 14PCh. 11 - Prob. 15PCh. 11 - Prob. 16PCh. 11 - Prob. 17PCh. 11 - 18. The speed of sound in air at room temperature...Ch. 11 - Prob. 19PCh. 11 - Prob. 20PCh. 11 - 21. A fisherman notices a buoy bobbing up and down...Ch. 11 - Prob. 22PCh. 11 - Prob. 23PCh. 11 - Prob. 24PCh. 11 - Prob. 25PCh. 11 - Prob. 26PCh. 11 - Prob. 27PCh. 11 - Prob. 28PCh. 11 - Prob. 29PCh. 11 - Problems 30–32. The graphs show displacement y as...Ch. 11 - Prob. 31PCh. 11 - 32. Rank the waves in order of maximum transverse...Ch. 11 - Prob. 33PCh. 11 - Prob. 34PCh. 11 - Prob. 35PCh. 11 - Prob. 36PCh. 11 - Prob. 37PCh. 11 - Prob. 38PCh. 11 - Prob. 39PCh. 11 - Prob. 40PCh. 11 - Prob. 41PCh. 11 - 42. A traveling sine wave is the result of the...Ch. 11 - Prob. 43PCh. 11 - Prob. 44PCh. 11 - Prob. 45PCh. 11 - Prob. 46PCh. 11 - Prob. 47PCh. 11 - Prob. 48PCh. 11 - Prob. 49PCh. 11 - Prob. 50PCh. 11 - Prob. 51PCh. 11 - Prob. 52PCh. 11 - Prob. 53PCh. 11 - Prob. 54PCh. 11 - Prob. 55PCh. 11 - Prob. 56PCh. 11 - Prob. 57PCh. 11 - Prob. 58PCh. 11 - Prob. 59PCh. 11 - Prob. 60PCh. 11 - Prob. 61PCh. 11 - Prob. 62PCh. 11 - Prob. 63PCh. 11 - Prob. 64PCh. 11 - Prob. 65PCh. 11 - Prob. 66PCh. 11 - Prob. 67PCh. 11 - Prob. 68PCh. 11 - Prob. 69PCh. 11 - Prob. 70PCh. 11 - Prob. 71PCh. 11 - Prob. 72PCh. 11 - Prob. 73PCh. 11 - Prob. 74PCh. 11 - Prob. 75PCh. 11 - Prob. 76PCh. 11 - Prob. 77PCh. 11 - Prob. 79PCh. 11 - Prob. 78PCh. 11 - Prob. 80PCh. 11 - Prob. 81PCh. 11 - Prob. 82PCh. 11 - Prob. 83PCh. 11 - Prob. 84PCh. 11 - Prob. 85PCh. 11 - Prob. 86PCh. 11 - Prob. 87PCh. 11 - Prob. 88PCh. 11 - Prob. 89PCh. 11 - Prob. 90PCh. 11 - Prob. 91PCh. 11 - Prob. 92PCh. 11 - Prob. 93PCh. 11 - Prob. 94PCh. 11 - Prob. 95PCh. 11 - Prob. 96PCh. 11 - Prob. 97PCh. 11 - Prob. 98PCh. 11 - Prob. 99PCh. 11 - Prob. 100PCh. 11 - Prob. 102PCh. 11 - 102. A harpsichord string is made of yellow brass...
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