College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Question
Chapter 14, Problem 48P

(a)

To determine

The expression for the frequency of standing wave.

(a)

Expert Solution
Check Mark

Answer to Problem 48P

Frequency of the standing wave is, fA=nA2LATAμA

Explanation of Solution

Given Info: Standing wave is setup in a string with both ends fixed.

Formula to calculate wave length of a wave produced in a string with both ends fixed is,

λA=2LAnA

  • λA is wave length of a wave produced in a string with both ends fixed.
  • LA is the length of the string.
  • nA is the number of antinodes.

For a string closed at both ends and the string vibrating at fundamental frequency one anti node is present at the centre of the standing wave and two nodes at the fixed position. So, the length of the string corresponds to half the wave length λ/2 of the standing wave. For each increase in antinode, half of the wave length of the wave is increased. For example, 1 antinode is present in λ/2 , 2 antinodes are present in λ/2+λ/2=2(λ/2) , 3 antinodes are present it λ/2+λ/2+λ/2=3(λ/2) and so on.

Formula to calculate the speed of the wave in a string is,

vA=TAμA

  • vA is velocity of the wave in a string.
  • TA is the tension in the string.
  • μA is the linear density of the string.

Conclusion:

Formula to calculate the frequency of a wave in terms of speed of the wave and wave length is,

fA=vAλA

  • fA is the frequency of the wave.

Substitute 2LA/nA for λA , TA/μA for vA to find fA .

fA=TA/μA2LA/nA=nA2LATAμA (I)

Thus, the expression for the frequency of a standing wave is fA=nA2LATAμA .

(b)

To determine

The expression for the frequency of standing wave with length of the string doubled.

(b)

Expert Solution
Check Mark

Answer to Problem 48P

Frequency of the standing wave with length doubled is fB=fA2 .

Explanation of Solution

Given Info: Length of the string is doubled LB=2LA . Number of antinodes is same as previous nB=nA . Tension in the string remains fixed TB=TA . Linear density of the wire remains fixed μB=μA .

From (a), formula to calculate the frequency of the wave in a string is,

fB=nB2LBTBμB (II)

  • fB is the frequency of the wave in a string with length doubled.
  • TA is the tension in the string.
  • μA is the linear density of the string.
  • LB is the length of the string.
  • nB is the number of antinodes.

Substitute nA for nB , 2LA for LB , TA for TB , and μA for μB to find fB .

fB=nA2(2LA)TAμA=fA2

Conclusion:

When the length of the string is doubled keeping all the other factors constant, the frequency of vibration of the string is reduced by half. Thus, the frequency of the wave in the string is fB=fA2 .

(c)

To determine

The tension that will produce nA+1 antinodes.

(c)

Expert Solution
Check Mark

Answer to Problem 48P

Tension in the string must be (nAnA+1)2TA .

Explanation of Solution

Given info: Frequency and length are held constant (fB=fA),(LB=LA) , Tension is altered (TBTA) , number of antinodes is increased by one nB=nA+1 .

Rewrite equation (II).

fB=nB2LBTBμB

Rearrange in terms of TB .

TB=4μBfB2LB2nB2 (III)

Substitute nA+1 for nB , fA for fB , LA for LB , and μA for μB to find TB .

TB=4μAfA2LA2(nA+1)2 (IV)

Conclusion:

Rearrange equation (I) in terms of TA .

TA=4μAfA2LA2nA2

Multiply and divide equation (IV) by nA2 .

TB=nA2(nA+1)24μAfA2LA2nA2=nA2(nA+1)2TA=(nAnA+1)2TA

Thus, the tension that will produce nA+1 antinodes is TB=(nAnA+1)2TA

(d)

To determine

Factor the tension should be changed to produce twice as many anti-nodes.

(d)

Expert Solution
Check Mark

Answer to Problem 48P

The factor the tension should be changed is TB=916TA .

Explanation of Solution

Given Info: Frequency is tripled fB=3fA , length of the string is halved LB=LA/2 , and antinodes are twice as before is nB=2nA .

Formula to calculate the tension from equation (III) is,

TB=4μBfB2LB2nB2

Substitute 2nA for nB , 3fA for fB , LA/2 for LB , and μA for μB to find TB .

TB=4μA(3fA)2(LA/2)2(2nA)2=9(4μAfA2LA2)16nA2=916TA

Conclusion:

When the frequency is tripled and length of the string is halved, then to produce twice as many antinodes than before tension in the string has to be reduced by a factor TB=916TA .

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