Chapter 32, Problem 39AP

Marie Cornu, a physicist at the Polytechnic Institute in Paris, invented phasors in about 1880. This problem helps you see their general utility in representing oscillations. Two mechanical vibrations are represented by the expressions

$y_1=12.0\sin 4.50t$

and

$y_2=12.0\sin (4.50t+70.0°)$

where y₁ and y₂ are in centimeters and t is in seconds. Find the amplitude and phase constant of the sum of these functions (a) by using a trigonometric identity (as from Appendix B) and (b) by representing the oscillations as phasors. (c) State the result of comparing the answers to parts (a) and (b). (d) Phasors make it equally easy to add traveling waves. Find the amplitude and phase constant of the sum of the three waves represented by

$\begin{array}{lll}{y}_1\hfill & =\hfill & 12.0\sin (15.0x-4.50t+70.0°)\hfill \\ y_2\hfill & =\hfill & 15.5\sin (15.0x-4.50t-80.0°)\hfill \\ y_3\hfill & =\hfill & 17.0\sin (15.0x-4.50t+160°)\hfill \end{array}$

where x, y_1, y₂, and y₃, are in centimeters and t is in seconds.

To determine

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

Answer to Problem 39AP

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

Explanation of Solution

Given info: The mechanical vibration of first wave is $y_1=12.0\sin 4.50t$ and for second wave is $y_2=12.0\sin \left(4.50t+70.0°\right)$.

Write the expression for the sum of two wave functions.

$y=y_1+y_2$

Here,

$y$ is the sum of two mechanical vibration.

$y_1$ is the mechanical vibration of first wave.

$y_2$ is the mechanical vibration of second wave.

Substitute $12.0\sin 4.50t$ for $y_1$ and $12.0\sin \left(4.50t+70.0°\right)$ for $y_2$.

$\begin{array}{c}y=12.0\sin 4.50t+12.0\sin \left(4.50t+70.0°\right)\\ =12.0\left[\sin 4.50t+\sin \left(4.50t+70.0°\right)\right]\\ =12.0\left[2\sin \left(\frac{4.50t+4.50t+70.0°}{2}\right)\cos \left(\frac{4.50t-\left(4.50t+70.0°\right)}{2}\right)\right]\\ =24.0\left[\sin \left(4.50t+35.0°\right)\cos \left(-35.0\right)\right]\end{array}$

Further solve the equation,

$y=19.7\sin \left(4.50t+35.0°\right)$

Conclusion:

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

To determine

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

Answer to Problem 39AP

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

Explanation of Solution

Given info: The mechanical vibration of first wave is $y_1=12.0\sin 4.50t$ and for second wave is $y_2=12.0\sin \left(4.50t+70.0°\right)$.

Write the expression for the phasor of a first oscillation.

$\overrightarrow{{\text{y}}_{\text{1}}}=\left(12.0\text{cm}\right)\widehat{\text{i}}$

Write the expression for the phasor of a second oscillation.

$\begin{array}{c}\overrightarrow{{\text{y}}_{\text{2}}}=\left(12.0\text{cm}\right)\left(\cos 70.0°\widehat{\text{i}}+\sin 70.0°\widehat{\text{j}}\right)\\ =\left(4.10\text{cm}\right)\widehat{\text{i}}+\left(11.27\text{cm}\right)\widehat{\text{j}}\end{array}$

Write the expression for the sum of two wave functions.

$\overrightarrow{\text{y}}=\overrightarrow{{\text{y}}_{\text{1}}}+\overrightarrow{{\text{y}}_{\text{2}}}$

Substitute $\left(12.0\text{cm}\right)\widehat{\text{i}}$ for $\overrightarrow{{\text{y}}_{\text{1}}}$ and $\left(4.10\text{cm}\right)\widehat{\text{i}}+\left(11.27\text{cm}\right)\widehat{\text{j}}$ for $\overrightarrow{{\text{y}}_{\text{2}}}$.

$\begin{array}{c}\overrightarrow{\text{y}}=\left(12.0\text{cm}\widehat{\text{i}}\right)+\left(\left(4.10\text{cm}\right)\widehat{\text{i}}+\left(11.27\text{cm}\right)\widehat{\text{j}}\right)\\ =\left(16.10\text{cm}\right)\widehat{\text{i}}+\left(11.27\text{cm}\right)\widehat{\text{j}}\end{array}$

Thus, the phasor representation of the sum of two wave functions is $\left(16.10\text{cm}\right)\widehat{\text{i}}+\left(11.27\text{cm}\right)\widehat{\text{j}}$.

Formula to calculate the amplitude of the resultant wave is,

$A=\sqrt{{\left(A_{\text{x}}\right)}^2+{\left(A_{\text{y}}\right)}^2}$

Here,

$A$ is the amplitude of the resultant wave.

$A_{\text{x}}$ is the amplitude of wave in $x$-direction.

$A_{\text{y}}$ is the amplitude of wave in $y$-direction.

Substitute $16.10\text{cm}$ for $A_{\text{x}}$ and $11.27\text{cm}$ for $A_{\text{y}}$ to find $A$.

$\begin{array}{c}A=\sqrt{{\left(16.10\text{cm}\right)}^2+{\left(11.27\text{cm}\right)}^2}\\ =19.7\text{cm}\end{array}$

Thus, the amplitude of the resultant wave is $19.7\text{cm}$.

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

$\tan \theta =\frac{A_{\text{y}}}{A_{\text{x}}}$

Substitute $16.10\text{cm}$ for $A_{\text{x}}$ and $11.27\text{cm}$ for $A_{\text{y}}$ to find $A$.

$\begin{array}{c}\tan \theta =\frac{11.27\text{cm}}{16.10\text{cm}}\\ \theta ={\tan }^{-1}\left(0.7\right)\\ =35.0°\end{array}$

Thus, phase difference between the resultant and the

Conclusion:

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

To determine

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

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 $y_1=12.0\sin 4.50t$ and for second wave is $y_2=12.0\sin \left(4.50t+70.0°\right)$.

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.

To determine

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

Answer to Problem 39AP

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

Explanation of Solution

Given info: The mechanical vibration of first wave is $y_1=12.0\sin \left(15.0x-4.50t+70.0°\right)$, for second wave is $y_2=15.5\sin \left(15.0x-4.50t-80.0°\right)$ and for second wave is $y_3=17.0\sin \left(15.0x-4.50t+160°\right)$.

Write the expression for the phasor of a first oscillation.

$\begin{array}{c}\overrightarrow{{\text{y}}_{\text{1}}}=\left(12.0\text{cm}\right)\left(\cos 70.0°\widehat{\text{i}}+\sin 70.0°\widehat{\text{j}}\right)\\ =\left(4.10\text{cm}\right)\widehat{\text{i}}+\left(11.27\text{cm}\right)\widehat{\text{j}}\end{array}$

Write the expression for the phasor of a second oscillation.

$\begin{array}{c}\overrightarrow{{\text{y}}_{\text{2}}}=\left(15.5\text{cm}\right)\left(\cos \left(-80.0°\right)\widehat{\text{i}}+\sin \left(-80.0°\right)\widehat{\text{j}}\right)\\ =\left(2.7\text{cm}\right)\widehat{\text{i}}-\left(15.26\text{cm}\right)\widehat{\text{j}}\end{array}$

Write the expression for the phasor of a third oscillation.

$\begin{array}{c}\overrightarrow{{\text{y}}_{\text{3}}}=\left(17.0\text{cm}\right)\left(\cos \left(160°\right)\widehat{\text{i}}+\sin \left(160°\right)\widehat{\text{j}}\right)\\ =\left(-15.97\text{cm}\right)\widehat{\text{i}}+\left(5.81\text{cm}\right)\widehat{\text{j}}\end{array}$

Write the expression for the sum of two wave functions.

$\overrightarrow{\text{y}}=\overrightarrow{{\text{y}}_{\text{1}}}+\overrightarrow{{\text{y}}_{\text{2}}}+\overrightarrow{{\text{y}}_{\text{3}}}$

Substitute $\left(4.10\text{cm}\right)\widehat{\text{i}}+\left(11.27\text{cm}\right)\widehat{\text{j}}$ for $\overrightarrow{{\text{y}}_{\text{1}}}$, $\left(2.7\text{cm}\right)\widehat{\text{i}}-\left(15.26\text{cm}\right)\widehat{\text{j}}$ for $\overrightarrow{{\text{y}}_{\text{2}}}$ and $\left(-15.97\text{cm}\right)\widehat{\text{i}}+\left(5.81\text{cm}\right)\widehat{\text{j}}$ for $\overrightarrow{{\text{y}}_{\text{3}}}$.

$\begin{array}{c}\overrightarrow{\text{y}}=\left[\begin{array}{l}\left(4.10\text{cm}\right)\widehat{\text{i}}+\left(11.27\text{cm}\right)\widehat{\text{j}}+\left(2.7\text{cm}\right)\widehat{\text{i}}\\ -\left(15.26\text{cm}\right)\widehat{\text{j}}+\left(-15.97\text{cm}\right)\widehat{\text{i}}+\left(5.81\text{cm}\right)\widehat{\text{j}}\end{array}\right]\\ =\left(-9.18\text{cm}\right)\widehat{\text{i}}+\left(1.83\text{cm}\right)\widehat{\text{j}}\end{array}$

Thus, the phasor representation of the sum of three wave functions is $\left(-9.18\text{cm}\right)\widehat{\text{i}}+\left(1.83\text{cm}\right)\widehat{\text{j}}$.

Formula to calculate the amplitude of the resultant wave is,

$A=\sqrt{{\left(A_{\text{x}}\right)}^2+{\left(A_{\text{y}}\right)}^2}$

Here,

$A$ is the amplitude of the resultant wave.

$A_{\text{x}}$ is the amplitude of wave in $x$-direction.

$A_{\text{y}}$ is the amplitude of wave in $y$-direction.

Substitute $-9.18\text{cm}$ for $A_{\text{x}}$ and $1.83\text{cm}$ for $A_{\text{y}}$ to find $A$.

$\begin{array}{c}A=\sqrt{{\left(-9.18\text{cm}\right)}^2+{\left(1.83\text{cm}\right)}^2}\\ =9.36\text{cm}\end{array}$

Thus, the amplitude of the resultant wave is $9.36\text{cm}$.

Formula to calculate the angle of the resultant wave is,

$\tan \theta =\frac{A_{\text{y}}}{A_{\text{x}}}$

Substitute $-9.18\text{cm}$ for $A_{\text{x}}$ and $1.83\text{cm}$ for $A_{\text{y}}$ to find $A$.

$\begin{array}{c}\tan \theta =\frac{1.83\text{cm}}{-9.18\text{cm}}\\ \theta ={\tan }^{-1}\left(-0.199\right)\\ =11.3°\end{array}$

Write the expression for the angle with the first wave.

$\theta \text{'}=180°-\theta$

Substitute $11.3°$ for $\theta$ to find $\theta \text{'}$.

$\begin{array}{c}\theta \text{'}=180°-11.3°\\ =168.7°\\ \approx 169°\end{array}$

Conclusion:

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