The expression for the frequency of standing wave.

Question

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Answer 1

Question

Chapter 14, Problem 48P

(a)

To determine

The expression for the frequency of standing wave.

(a)

Expert Solution

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

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

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 $(TB≠TA)$ , 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

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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Answer 2

Question

Chapter 14, Problem 48P

(a)

To determine

The expression for the frequency of standing wave.

(a)

Expert Solution

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

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

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 $(TB≠TA)$ , 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

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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Answer 3

Question

Chapter 14, Problem 48P

(a)

To determine

The expression for the frequency of standing wave.

(a)

Expert Solution

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

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

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 $(TB≠TA)$ , 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

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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Answer 4

Question

Chapter 14, Problem 48P

(a)

To determine

The expression for the frequency of standing wave.

(a)

Expert Solution

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

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

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 $(TB≠TA)$ , 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

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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Answer 5

Chapter 14, Problem 48P

(a)

To determine

The expression for the frequency of standing wave.

(a)

Expert Solution

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

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

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 $(TB≠TA)$ , 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

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 $TB=916TA$ .

Answer 6

Chapter 14, Problem 48P

(a)

To determine

The expression for the frequency of standing wave.

(a)

Expert Solution

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 $fA=nA2LATAμA$ .

Answer 7

Chapter 14, Problem 48P

(a)

To determine

The expression for the frequency of standing wave.

Answer 8

(a)

Expert Solution

Answer to Problem 48P

Frequency of the standing wave is,

$fA=nA2LATAμA$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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 $fA=nA2LATAμA$ .

Answer 13

(b)

To determine

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

(b)

Expert Solution

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 $fB=fA2$ .

Answer 14

(b)

To determine

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

Answer 15

(b)

Expert Solution

Answer to Problem 48P

Frequency of the standing wave with length doubled is

$fB=fA2$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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 $fB=fA2$ .

Answer 20

(c)

To determine

The tension that will produce $nA+1$ antinodes.

(c)

Expert Solution

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 $(TB≠TA)$ , 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 $TB=(nAnA+1)2TA$

Answer 21

(c)

To determine

The tension that will produce $nA+1$ antinodes.

Answer 22

(c)

Expert Solution

Answer to Problem 48P

Tension in the string must be

$(nAnA+1)2TA$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: Frequency and length are held constant $(fB=fA),(LB=LA)$ , Tension is altered $(TB≠TA)$ , 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 $TB=(nAnA+1)2TA$

Answer 27

(d)

To determine

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

(d)

Expert Solution

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 $TB=916TA$ .

Answer 28

(d)

To determine

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

Answer 29

(d)

Expert Solution

Answer to Problem 48P

The factor the tension should be changed is

$TB=916TA$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

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 $TB=916TA$ .