EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 16, Problem 59P

(a)

To determine

The fundamental frequency of the string.

(a)

Expert Solution
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Explanation of Solution

Given:

Linear mass density of the string is 4.00×103kg/m .

Tension in the string is 360N .

One of the resonance frequency is 375Hz .

Other higher resonance frequency is 450Hz .

Formula used:

Write expression for nth multiple of fundamental frequency.

  nf1=fn

Here, f1 is the fundamental frequency and fn is the nth multiple of fundamental frequency.

Rearrange above expression for n .

  n=fnf1

Write expression for (n+1)th multiple of fundamental frequency.

  (n+1)f1=fn+1

Here, fn+1 is the (n+1)th multiple of fundamental frequency.

Substitute fnf1 for n in above expression and solve.

  fn+f1=fn+1......... (1)

Calculation:

Substitute 375Hz for fn and 450Hz for fn+1 in equation (1) and solve for f1 .

  375Hz+f1=450Hzf1=75Hz

Conclusion:

Thus, fundamental frequency of the string is 75Hz .

(b)

To determine

The harmonic which has given frequencies.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Linear mass density of the string is 4.00×103kg/m .

Tension in the string is 360N .

One of the resonance frequency is 375Hz .

Other higher resonance frequency is 450Hz .

Formula used:

Write expression for nth multiple of fundamental frequency.

  nf1=fn......... (1)

Here, f1 is the fundamental frequency and fn is the nth multiple of fundamental frequency.

Rearrange above expression for n .

  n=fnf1

Write expression for (n+1)th multiple of fundamental frequency.

  (n+1)f1=fn+1

Here, fn+1 is the (n+1)th multiple of fundamental frequency.

Substitute fnf1 for n in above expression and solve.

  fn+f1=fn+1......... (2)

Calculation:

Substitute 375Hz for fn and 450Hz for fn+1 in equation (2) and solve for f1 .

  375Hz+f1=450Hzf1=75Hz

Substitute 75Hz for f1 and 375Hz for fn in equation (1).

  n(75Hz)=375Hzn=5

Conclusion:

Thus, fifth harmonic has the given frequencies.

(c)

To determine

The length of the string.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

Linear mass density of the string is 4.00×103kg/m .

Tension in the string is 360N .

One of the resonance frequency is 375Hz .

Other higher resonance frequency is 450Hz .

Formula used:

Write expression for nth multiple of fundamental frequency.

  nf1=fn

Here, f1 is the fundamental frequency and fn is the nth multiple of fundamental frequency.

Rearrange above expression for n .

  n=fnf1

Write expression for (n+1)th multiple of fundamental frequency.

  (n+1)f1=fn+1

Here, fn+1 is the (n+1)th multiple of fundamental frequency.

Substitute fnf1 for n in above expression and solve.

  fn+f1=fn+1......... (1)

Write expression for length of the string.

  L=12f1Tμ......... (2)

Here, T is the tension in the string and μ is the linear mass density of the string.

Calculation:

Substitute 375Hz for fn and 450Hz for fn+1 in equation (1) and solve for f1 .

  375Hz+f1=450Hzf1=75Hz

Substitute 75Hz for f1 , 360N for T and 4.00×103kg/m for μ in equation (2).

  L=12( 75Hz) ( 360N ) 4.00× 10 3 kg/m =2m

Conclusion:

Thus, the length of the string is 2m .

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