Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 15, Problem 47P

(a)

To determine

To calculate:

Calculate the pressure amplitude of the wave.

(a)

Expert Solution
Check Mark

Answer to Problem 47P

The pressure amplitude (p0) of the wave is 0.75Pa .

Explanation of Solution

Given:

Pressure variation is given by,

  p(x,t)=0.75cos[π2(x-343t)]

Formula used:

The pressure amplitude can be calculated by using:

  p(x,t)=0.75cos[π2(x-343t)]

Where,

  k= Wave number

  p0= Pressure amplitude,

  v= Wave speed

The pressure variation is given as, p(x,t)=0.75cos[π2(x-343t)]

Where,

  k= Wave number

  p0= Pressure amplitude,

  v= Wave speed

From the inspection of the equation which is given,

  p(x,t)=0.75cos[π2(x-343t)]

The value of the p0=0.75Pa

Conclusion:

Thus, the pressure amplitude (p0) of the wave is 0.75Pa .

(b)

To determine

To calculate:

Calculate the wavelength of the wave.

(b)

Expert Solution
Check Mark

Answer to Problem 47P

The wavelength (λ) of the wave is 4.00m .

Explanation of Solution

Given:

Pressure variation is given by,

  p(x,t)=0.75cos[π2(x-343t)] .

Formula used:

The wavelength can be calculated by using:

  p(x,t)=0.75cos[π2(x-343t)]

Where,

  k= Wave number

  p0= Pressure amplitude

  v= Wave speed

The pressure variation is given as,

  p(x,t)=0.75cos[π2(x-343t)]

Where,

  k= Wave number

  p0= Pressure amplitude

  v= Wave speed

As the value of the k ,

  k=λ=π2

Thus, λ=4.00m

Conclusion:

Thus, the wavelength (λ) of the wave is 4.00m .

(c)

To determine

To calculate:

Calculate the frequency of the wave.

(c)

Expert Solution
Check Mark

Answer to Problem 47P

The frequency (f) of the wave is 85.8Hz .

Explanation of Solution

Given:

Pressure variation is given by,

  p(x,t)=0.75cos[π2(x-343t)] .

Formula used:

Frequency,

  f=kv

Where,

  f= Frequency of the wave

  v= Wave speed

  k= Wave number

The pressure variation is given as,

  p(x,t)=0.75cos[π2(x-343t)]

Where,

  k= Wave number

  p0= Pressure amplitude

  v= Wave speed

Solve

  v=ωk=2πfk

To get frequency,

  f=kv

Substitute the numerical values in the above equation,

  f=kvf=π2( 343m/s)f=85.8Hz

Conclusion:

Thus, the frequency (f) of the wave is 85.8Hz .

(d)

To determine

To calculate:

Calculate the speed of the wave.

(d)

Expert Solution
Check Mark

Answer to Problem 47P

The speed (v) of the wave is 343m/s .

Explanation of Solution

Given:

Pressure variation is given by,

  p(x,t)=0.75cos[π2(x-343t)]

Formula used:

The pressure variation is given as,

  p(x,t)=0.75cos[π2(x-343t)]

Where,

  k= Wave number

  p0= Pressure amplitude

  v= Wave speed

The pressure variation is given as,

  p(x,t)=0.75cos[π2(x-343t)]

From the inspection of the equation which is given,

  p(x,t)=0.75cos[π2(x-343t)]

The value of v=343m/s .

Conclusion:

Thus, the speed (v) of wave is 343m/s .s

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The pressure in a traveling sound wave is given by the equationΔP = (1.62 Pa) sin [π (0.714 m-1)x - π (422 s-1)t]Find (a) the pressure amplitude, (b) the frequency, (c) the wavelength, and (d) the speed of the wave .
(a) An experimenter wishes to generate in air a sound wave that has a displacement amplitude of 6.20   10-6 m. The pressure amplitude is to be limited to0.850 Pa. What is the minimum wavelength the sound wave can have? (Take the equilibrium density of air to be ρ = 1.20 kg/m3 and assume the speed of sound in air is v = 343 m/s.)   (b) Calculate the pressure amplitude of a 2.80 kHz sound wave in air, assuming that the displacement amplitude is equal to 2.00 ✕ 10-8 m.[Note: Use the following values, as needed. The equilibrium density of air is ρ = 1.20 kg/m3. The speed of sound in air is v = 343 m/s. Pressure variations ΔP are measured relative to atmospheric pressure, 1.013 ✕ 105 Pa.]   (c) Earthquakes at fault lines in Earth's crust create seismic waves, which are longitudinal (P-waves) or transverse (S-waves). The P-waves have a speed of about 9 km/s. Estimate the average bulk modulus of Earth's crust given that the density of rock is about 2500 kg/m3.
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Chapter 15 Solutions

Physics for Scientists and Engineers

Ch. 15 - Prob. 11PCh. 15 - Prob. 12PCh. 15 - Prob. 13PCh. 15 - Prob. 14PCh. 15 - Prob. 15PCh. 15 - Prob. 16PCh. 15 - Prob. 17PCh. 15 - Prob. 18PCh. 15 - Prob. 19PCh. 15 - Prob. 20PCh. 15 - Prob. 21PCh. 15 - Prob. 22PCh. 15 - Prob. 23PCh. 15 - Prob. 24PCh. 15 - Prob. 25PCh. 15 - Prob. 26PCh. 15 - Prob. 27PCh. 15 - Prob. 28PCh. 15 - Prob. 29PCh. 15 - Prob. 30PCh. 15 - Prob. 31PCh. 15 - Prob. 32PCh. 15 - Prob. 33PCh. 15 - Prob. 34PCh. 15 - Prob. 35PCh. 15 - Prob. 36PCh. 15 - Prob. 37PCh. 15 - Prob. 38PCh. 15 - Prob. 39PCh. 15 - Prob. 40PCh. 15 - Prob. 41PCh. 15 - Prob. 42PCh. 15 - Prob. 43PCh. 15 - Prob. 44PCh. 15 - Prob. 45PCh. 15 - Prob. 46PCh. 15 - Prob. 47PCh. 15 - Prob. 48PCh. 15 - Prob. 49PCh. 15 - Prob. 50PCh. 15 - Prob. 51PCh. 15 - Prob. 52PCh. 15 - Prob. 53PCh. 15 - Prob. 54PCh. 15 - Prob. 55PCh. 15 - Prob. 56PCh. 15 - Prob. 57PCh. 15 - Prob. 58PCh. 15 - Prob. 59PCh. 15 - Prob. 60PCh. 15 - Prob. 61PCh. 15 - Prob. 62PCh. 15 - Prob. 63PCh. 15 - Prob. 64PCh. 15 - Prob. 65PCh. 15 - Prob. 66PCh. 15 - Prob. 67PCh. 15 - Prob. 68PCh. 15 - Prob. 69PCh. 15 - Prob. 70PCh. 15 - Prob. 71PCh. 15 - Prob. 72PCh. 15 - Prob. 73PCh. 15 - Prob. 74PCh. 15 - Prob. 75PCh. 15 - Prob. 76PCh. 15 - Prob. 77PCh. 15 - Prob. 78PCh. 15 - Prob. 79PCh. 15 - Prob. 80PCh. 15 - Prob. 81PCh. 15 - Prob. 82PCh. 15 - Prob. 83PCh. 15 - Prob. 84PCh. 15 - Prob. 85PCh. 15 - Prob. 86PCh. 15 - Prob. 87PCh. 15 - Prob. 88PCh. 15 - Prob. 89PCh. 15 - Prob. 90PCh. 15 - Prob. 91PCh. 15 - Prob. 92PCh. 15 - Prob. 93PCh. 15 - Prob. 94PCh. 15 - Prob. 95PCh. 15 - Prob. 96PCh. 15 - Prob. 97PCh. 15 - Prob. 98PCh. 15 - Prob. 99PCh. 15 - Prob. 100PCh. 15 - Prob. 101PCh. 15 - Prob. 102PCh. 15 - Prob. 103PCh. 15 - Prob. 104P
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