Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Question
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Chapter 16, Problem 76P

(a)

To determine

The two resonant frequencies.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The distance between nodes for first case is 6.94cm .

The distance between nodes for second case is 5.40cm .

Formula used:

Write expression for the wavelength of the wave.

  λn=2d

Here, d is the distance between two nodes.

Write expression for frequency.

  fn=vλn

Here, v is the velocity of the sound.

Substitute 2d for λn in above expression.

  fn=v2d........ (1)

Calculation:

Substitute 343m/s for v and 6.94cm for d in equation (1).

  fn=343m/s2( 6.94cm)( 1m 100cm )fn=2471Hz

Substitute fn

  343m/s for v and 5.40cm for d in equation (1).

  fn=343m/s2( 5.40cm)( 1m 100cm )fn=3175Hz

Conclusion:

Thus, the two frequencies are 2471Hz and 3175Hz .

(b)

To determine

The fundamental frequency of the tube.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The distance between nodes for first case is 6.94cm .

The distance between nodes for second case is 5.40cm .

Formula used:

Write expression for the wavelength of the wave.

  λn=2d

Here, d is the distance between two nodes.

Write expression for frequency.

  fn=vλn

Here, v is the velocity of the sound.

Substitute 2d for λn in above expression.

  fn=v2d........ (1)

Write expression for frequency of nth resonance.

  fn=nf1

Rearrange above expression for n .

  n=fnf1

Write expression for next resonance.

  fn+2=(n+2)f1

Substitute fnf1 for n in above expression.

  fn+2=(fnf1+2)f1........ (2)

Calculation:

Substitute 343m/s for v and 6.94cm for d in equation (1).

  fn=343m/s2( 6.94cm)( 1m 100cm )fn=2471Hz

Substitute fn+2 for fn , 343m/s for v and 5.40cm for d in equation (1).

  fn+2=343m/s2( 5.40cm)( 1m 100cm )fn+2=3175Hz

Substitute 2471Hz for fn and 3175Hz for fn+2 and solve for f1 in equation (2).

  3175Hz=( 2471Hz f 1 +2)f1f1=353Hz

Conclusion:

Thus, the fundamental frequency is 353Hz .

(c)

To determine

The value of n and n+2 .

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The distance between nodes for first case is 6.94cm .

The distance between nodes for second case is 5.40cm .

Formula used:

Write expression for the wavelength of the wave.

  λn=2d

Here, d is the distance between two nodes.

Write expression for frequency.

  fn=vλn

Here, v is the velocity of the sound.

Substitute 2d for λn in above expression.

  fn=v2d........ (1)

Write expression for frequency of nth resonance.

  fn=nf1

Rearrange above expression for n .

  n=fnf1........ (2)

Write expression for next resonance.

  fn+2=(n+2)f1

Rearrange above expression for n+2 .

  n+2=fn+2fn........ (3)

Substitute fnf1 for n in above expression.

  fn+2=(fnf1+2)f1........ (4)

Calculation:

Substitute 343m/s for v and 6.94cm for d in equation (1).

  fn=343m/s2( 6.94cm)( 1m 100cm )fn=2471Hz

Substitute fn+2 for fn , 343m/s for v and 5.40cm for d in equation (1).

  fn+2=343m/s2( 5.40cm)( 1m 100cm )fn+2=3175Hz

Substitute 2471Hz for fn and 3175Hz for fn+2 and solve for f1 in equation (4).

  3175Hz=( 2471Hz f 1 +2)f1f1=353Hz

Substitute 353Hz for f1 and 2471Hz for fn in equation (2).

  n=2471Hz353Hzn=7

Substitute 353Hz for f1 and 3175Hz for fn+2 in equation (3).

  n+2=3175Hz353Hzn+2=9

Conclusion:

Thus, the value of n is 7 and n+2 is 9 .

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