COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
Question
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Chapter 13, Problem 45QAP
To determine

(a)

The value of amplitude of pressure wave.

Expert Solution
Check Mark

Answer to Problem 45QAP

  A=1.0atm

Explanation of Solution

Given:

  p(x,t)=(1.0atm)cos(6.0x4.0t)

Formula used:

  y(x,t)=Acos(kxωt)

Calculation:

The displacement of transverse wave is given by,

  p(x,t)=Acos(kxωt)

Comparing above equation with the give equation,

  p(x,t)=(1.0atm)cos(6.0x4.0t)

Amplitude, A= 0.5 atm

Conclusion:

Amplitude of the pressure wave is 0.5atm.

To determine

(b)

The value of wave number of given pressure wave.

Expert Solution
Check Mark

Answer to Problem 45QAP

  k=6.0rad/m

Explanation of Solution

Given:

  p(x,t)=(1.0atm)cos(6.0x4.0t)

Formula used:

  y(x,t)=Acos(kxωt)

Calculation:

The displacement of transverse wave is given by,

  p(x,t)=(1.0atm)cos(6.0x4.0t)

Comparing above equation with the give equation,

  y(x,t)=Acos(kxωt)

  k=6.0rad/m

Conclusion:

Wave number of the pressure wave is k=6.0rad/m

To determine

(c)

The value of frequency of pressure wave.

Expert Solution
Check Mark

Answer to Problem 45QAP

  f=0.6369Hz

Explanation of Solution

Given:

  p(x,t)=(1.0atm)cos(6.0x4.0t)

Formula used:

  y(x,t)=Acos(kxωt)

Calculation:

The displacement of transverse wave is given by,

  p(x,t)=(1.0atm)cos(6.0x4.0t)

Comparing above equation with the give equation,

  y(x,t)=Acos(kxωt)

We get,

  ω=4rad/s2πf=4f=23.14f=0.6369Hz

Conclusion:

Frequency of the pressure wave is f=0.6369Hz.

To determine

(d)

The value of speed of pressure wave.

Expert Solution
Check Mark

Answer to Problem 45QAP

  vp=0.7287m/s

Explanation of Solution

Given:

  p(x,t)=(1.0atm)cos(6.0x4.0t)

Formula used:

  y(x,t)=Acos(kxωt)

Calculation:

The displacement of transverse wave is given by,

  p(x,t)=(1.0atm)cos(6.0x4.0t)

Comparing above equation with the give equation,

  y(x,t)=Acos(kxωt)

We get,

  k=2πλ=6but,vp=λfλ=vpf6=2π v pf=2πfvp6=2πfvpvp=2π(0.6963)6vp=0.7287m/s

Conclusion:

Speed of the pressure wave is vp=0.7287m/s.

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Chapter 13 Solutions

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

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