Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 32, Problem 39AP

(a)

To determine

The amplitude and phase constant of the sum of the given function by using trigonometry identity.

(a)

Expert Solution
Check Mark

Answer to Problem 39AP

The amplitude of the sum of the given function by trigonometry identity is 19.7cm and a constant phase difference of 35.0° from the first wave.

Explanation of Solution

Given info: The mechanical vibration of first wave is y1=12.0sin4.50t and for second wave is y2=12.0sin(4.50t+70.0°) .

Write the expression for the sum of two wave functions.

y=y1+y2

Here,

y is the sum of two mechanical vibration.

y1 is the mechanical vibration of first wave.

y2 is the mechanical vibration of second wave.

Substitute 12.0sin4.50t for y1 and 12.0sin(4.50t+70.0°) for y2 .

y=12.0sin4.50t+12.0sin(4.50t+70.0°)=12.0[sin4.50t+sin(4.50t+70.0°)]=12.0[2sin(4.50t+4.50t+70.0°2)cos(4.50t(4.50t+70.0°)2)]=24.0[sin(4.50t+35.0°)cos(35.0)]

Further solve the equation,

y=19.7sin(4.50t+35.0°)

Conclusion:

Therefore, the amplitude of the sum of the given function by trigonometry identity is 19.7cm and a constant phase difference of 35.0° from the first wave.

(b)

To determine

The amplitude and phase constant of the sum of the given function by representing the oscillation as phasors.

(b)

Expert Solution
Check Mark

Answer to Problem 39AP

The amplitude of the sum of the given function by phasor representation is 19.7cm and a constant phase difference of 35.0° from the first wave.

Explanation of Solution

Given info: The mechanical vibration of first wave is y1=12.0sin4.50t and for second wave is y2=12.0sin(4.50t+70.0°) .

Write the expression for the phasor of a first oscillation.

y1=(12.0cm)i^

Write the expression for the phasor of a second oscillation.

y2=(12.0cm)(cos70.0°i^+sin70.0°j^)=(4.10cm)i^+(11.27cm)j^

Write the expression for the sum of two wave functions.

y=y1+y2

Substitute (12.0cm)i^ for y1 and (4.10cm)i^+(11.27cm)j^ for y2 .

y=(12.0cmi^)+((4.10cm)i^+(11.27cm)j^)=(16.10cm)i^+(11.27cm)j^

Thus, the phasor representation of the sum of two wave functions is (16.10cm)i^+(11.27cm)j^ .

Formula to calculate the amplitude of the resultant wave is,

A=(Ax)2+(Ay)2

Here,

A is the amplitude of the resultant wave.

Ax is the amplitude of wave in x -direction.

Ay is the amplitude of wave in y -direction.

Substitute 16.10cm for Ax and 11.27cm for Ay to find A .

A=(16.10cm)2+(11.27cm)2=19.7cm

Thus, the amplitude of the resultant wave is 19.7cm .

Formula to calculate the angle of the resultant wave makes with the first wave is,

tanθ=AyAx

Substitute 16.10cm for Ax and 11.27cm for Ay to find A .

tanθ=11.27cm16.10cmθ=tan1(0.7)=35.0°

Thus, phase difference between the resultant and the

Conclusion:

Therefore, the amplitude of the sum of the given function by phasor representation is 19.7cm and a constant phase difference of 35.0° from the first wave.

(c)

To determine

The result by compare the answer to part (a) and part (b).

(c)

Expert Solution
Check Mark

Answer to Problem 39AP

The result of part (a) and part (b) are identical.

Explanation of Solution

Given info: The mechanical vibration of first wave is y1=12.0sin4.50t and for second wave is y2=12.0sin(4.50t+70.0°) .

Since from the trigonometry identities the amplitude and the phase angle of the sum of two waves are identical to the amplitude and the phase angle of the sum of two waves by phasor representation, hence the both the method is valid to estimate the amplitude and the phase angle of the resultant wave.

Conclusion:

Therefore, the result of part (a) and part (b) are identical.

(d)

To determine

The amplitude and phase constant of the sum of the given function by represent the oscillation as phasors.

(d)

Expert Solution
Check Mark

Answer to Problem 39AP

The amplitude of the sum of the given function by phasor representation is 9.36cm and a constant phase difference of 169° from the first wave.

Explanation of Solution

Given info: The mechanical vibration of first wave is y1=12.0sin(15.0x4.50t+70.0°) , for second wave is y2=15.5sin(15.0x4.50t80.0°) and for second wave is y3=17.0sin(15.0x4.50t+160°) .

Write the expression for the phasor of a first oscillation.

y1=(12.0cm)(cos70.0°i^+sin70.0°j^)=(4.10cm)i^+(11.27cm)j^

Write the expression for the phasor of a second oscillation.

y2=(15.5cm)(cos(80.0°)i^+sin(80.0°)j^)=(2.7cm)i^(15.26cm)j^

Write the expression for the phasor of a third oscillation.

y3=(17.0cm)(cos(160°)i^+sin(160°)j^)=(15.97cm)i^+(5.81cm)j^

Write the expression for the sum of two wave functions.

y=y1+y2+y3

Substitute (4.10cm)i^+(11.27cm)j^ for y1 , (2.7cm)i^(15.26cm)j^ for y2 and (15.97cm)i^+(5.81cm)j^ for y3 .

y=[(4.10cm)i^+(11.27cm)j^+(2.7cm)i^(15.26cm)j^+(15.97cm)i^+(5.81cm)j^]=(9.18cm)i^+(1.83cm)j^

Thus, the phasor representation of the sum of three wave functions is (9.18cm)i^+(1.83cm)j^ .

Formula to calculate the amplitude of the resultant wave is,

A=(Ax)2+(Ay)2

Here,

A is the amplitude of the resultant wave.

Ax is the amplitude of wave in x -direction.

Ay is the amplitude of wave in y -direction.

Substitute 9.18cm for Ax and 1.83cm for Ay to find A .

A=(9.18cm)2+(1.83cm)2=9.36cm

Thus, the amplitude of the resultant wave is 9.36cm .

Formula to calculate the angle of the resultant wave is,

tanθ=AyAx

Substitute 9.18cm for Ax and 1.83cm for Ay to find A .

tanθ=1.83cm9.18cmθ=tan1(0.199)=11.3°

Write the expression for the angle with the first wave.

θ'=180°θ

Substitute 11.3° for θ to find θ' .

θ'=180°11.3°=168.7°169°

Conclusion:

Therefore, the amplitude of the sum of the given function by phasor representation is 9.36cm and a constant phase difference of 169° from the first wave.

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Chapter 32 Solutions

Physics for Scientists and Engineers with Modern Physics

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