Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 12, Problem 81P

(a)

To determine

The wavelength of the fundamental mode of vibration.

(a)

Expert Solution
Check Mark

Answer to Problem 81P

The wavelength of the fundamental mode is 1.28m.

Explanation of Solution

The length of the string is 64.0cm, the tension on the string is 133N and the fundamental frequency is 110.0Hz.

Write the expression for wavelength of the fundamental mode

λ=2L                                                            (I)

Here, λ is the wavelength of the fundamental mode and L is the length of the string.

Substitute 64.0cm for L in (I) to find λ

λ=2(64.0cm)=2(64.0×102m)=1.28m

Thus, the wavelength of the fundamental mode is 1.28m.

(b)

To determine

The speed of the wave.

(b)

Expert Solution
Check Mark

Answer to Problem 81P

The speed of the wave is 141ms1.

Explanation of Solution

Write the expression for speed of the wave

v=λf                                                            (II)

Here, v is the wave speed and f is the frequency.

Substitute 1.28m for λ and 110.0Hz for f in (II) to find v

v=(1.28m)(110.0Hz1s11Hz)=141ms1

Thus, the speed of the wave is 141ms1.

(c)

To determine

The linear mass density of the string.

(c)

Expert Solution
Check Mark

Answer to Problem 81P

The linear mass density of the string is 6.71 gm1.

Explanation of Solution

Write the expression for velocity of wave propagation on the string

v=Tμ                                                                   (III)

Here, T is the tension on the wire and μ is the linear mass density.


Rearrange for μ

μ=Tv2                                                                    (IV)

Substitute 141ms1 for v and 133N for T in (IV) to find μ

μ=133N(141ms1)2=(133kgs2)(141ms1)2=6.71×103kgm1=6.71 gm1

Thus, the linear mass density of the string is 6.71 gm1.

(d)

To determine

The maximum speed of a point on the vibrating string.

(d)

Expert Solution
Check Mark

Answer to Problem 81P

The maximum speed of a point on the vibrating string 1.59ms1.

Explanation of Solution

The maximum amplitude on the string is 2.30mm and the fundamental frequency is 110.0Hz.

Write the expression for maximum speed

vm=ωA                                                             (V)

Here, vm is the maximum speed of a point, ω is the angular frequency and A is the maximum amplitude.


Substitute for ω

vm=2πfA                                                                  (VI)

Here, f is the fundamental frequency.

Substitute 110.0Hz for f and 2.30mm for A in (VI) to find vm

vm=2π(110.0Hz)(2.30mm)=2π(110.0s1)(2.30×103m)=1588×103 ms11.59ms1

Thus, the linear mass density of the string is 1.59ms1.

(e)

To determine

The frequency of the sound wave.

(e)

Expert Solution
Check Mark

Answer to Problem 81P

The frequency of the sound wave in air is 110.0Hz.

Explanation of Solution

The frequency of the vibrating string is 110.0Hz.

The vibrating string puts the air molecules around it into vibratory motion. The frequency of vibration of the air molecules is same as the frequency of the vibrating string.

Hence the frequency of the sound wave in air is 110.0Hz.

(f)

To determine

The wavelength of sound in air medium at 20.0°C .

(f)

Expert Solution
Check Mark

Answer to Problem 81P

The wavelength of sound at 20.0° C is 3.120m.

Explanation of Solution

The air temperature in kelvin is (20.0+273.15)K.

Rearrange (II)

λ=vf                                                                  (VII)

Here, v is the velocity of sound in air, λ is the wavelength in air and f is the frequency in air.

Write the expression for temperature dependence of velocity of sound

v=v0θθ0                                                                (VIII)

Here, v0 is the velocity of sound at absolute temperature θ is the given temperature in kelvin and θ0 is the absolute temperature in kelvin

Substitute 343ms1 for v0, (20.0+273.15)K for θ , and 273.15K for θ0 in (VIII) to find v.

v=343ms1(20.0+273.15)K273.15K=343.2ms1

Substitute 343.2ms1 for v and 110.0Hz for f in (VII) to find λ

λ=343.2ms1110.0Hz=343.2ms1110.0Hz=3.120m

Thus, the wavelength of sound at 20.0° C is 3.120m.

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Chapter 12 Solutions

Physics

Ch. 12.8 - Prob. 12.8CPCh. 12.8 - Prob. 12.8PPCh. 12.8 - Prob. 12.9PPCh. 12 - Prob. 1CQCh. 12 - Prob. 2CQCh. 12 - Prob. 3CQCh. 12 - Prob. 4CQCh. 12 - Prob. 5CQCh. 12 - Prob. 6CQCh. 12 - Prob. 7CQCh. 12 - Prob. 8CQCh. 12 - Prob. 9CQCh. 12 - Prob. 10CQCh. 12 - Prob. 11CQCh. 12 - Prob. 12CQCh. 12 - Prob. 13CQCh. 12 - Prob. 14CQCh. 12 - Prob. 15CQCh. 12 - Prob. 16CQCh. 12 - Prob. 17CQCh. 12 - Prob. 1MCQCh. 12 - Prob. 2MCQCh. 12 - Prob. 3MCQCh. 12 - Prob. 4MCQCh. 12 - Prob. 5MCQCh. 12 - Prob. 6MCQCh. 12 - Prob. 7MCQCh. 12 - Prob. 8MCQCh. 12 - Prob. 9MCQCh. 12 - Prob. 10MCQCh. 12 - Prob. 11MCQCh. 12 - Prob. 1PCh. 12 - Prob. 2PCh. 12 - Prob. 3PCh. 12 - Prob. 4PCh. 12 - Prob. 5PCh. 12 - Prob. 6PCh. 12 - Prob. 7PCh. 12 - Prob. 8PCh. 12 - Prob. 9PCh. 12 - Prob. 10PCh. 12 - Prob. 11PCh. 12 - Prob. 12PCh. 12 - 13. Six sound waves have pressure amplitudes p0and...Ch. 12 - Prob. 14PCh. 12 - Prob. 15PCh. 12 - Prob. 16PCh. 12 - Prob. 17PCh. 12 - Prob. 18PCh. 12 - Prob. 19PCh. 12 - Prob. 20PCh. 12 - Prob. 21PCh. 12 - Prob. 22PCh. 12 - Prob. 23PCh. 12 - Prob. 24PCh. 12 - Prob. 25PCh. 12 - Prob. 26PCh. 12 - Prob. 27PCh. 12 - Prob. 28PCh. 12 - Prob. 29PCh. 12 - Prob. 30PCh. 12 - Prob. 31PCh. 12 - Prob. 32PCh. 12 - Prob. 33PCh. 12 - Prob. 34PCh. 12 - Prob. 35PCh. 12 - Prob. 36PCh. 12 - Prob. 37PCh. 12 - Prob. 38PCh. 12 - Prob. 39PCh. 12 - Prob. 40PCh. 12 - Prob. 41PCh. 12 - Prob. 42PCh. 12 - Prob. 43PCh. 12 - Prob. 44PCh. 12 - Prob. 45PCh. 12 - Prob. 46PCh. 12 - Prob. 47PCh. 12 - Prob. 48PCh. 12 - Prob. 49PCh. 12 - Prob. 50PCh. 12 - Prob. 51PCh. 12 - Prob. 52PCh. 12 - Prob. 53PCh. 12 - Prob. 54PCh. 12 - Prob. 55PCh. 12 - Prob. 56PCh. 12 - Prob. 57PCh. 12 - Prob. 58PCh. 12 - Prob. 59PCh. 12 - Prob. 60PCh. 12 - Prob. 61PCh. 12 - Prob. 62PCh. 12 - Prob. 63PCh. 12 - Prob. 64PCh. 12 - Prob. 65PCh. 12 - Prob. 66PCh. 12 - Prob. 67PCh. 12 - Prob. 68PCh. 12 - Prob. 69PCh. 12 - 70. Some bats determine their distance to an...Ch. 12 - Prob. 71PCh. 12 - Prob. 72PCh. 12 - Prob. 73PCh. 12 - Prob. 74PCh. 12 - Prob. 75PCh. 12 - Prob. 76PCh. 12 - Prob. 77PCh. 12 - Prob. 79PCh. 12 - Prob. 78PCh. 12 - Prob. 80PCh. 12 - Prob. 81PCh. 12 - Prob. 82P
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