Question

Three sinusoidal waves of the same frequency travel along a string in the positive direction of an xaxis. Their amplitudes are ${\mathbf{y}}_{\mathbf{1}}$,${\mathbf{y}}_{\mathbf{1}}\mathbf{/}\mathbf{2}$, and${\mathbf{y}}_{\mathbf{1}}\mathbf{/}\mathbf{3}$, and their phase constants are 0,$\mathbf{\pi }\mathbf{/}\mathbf{2}$, and$\mathbf{\pi }$, respectively. What are the (a) amplitude and (b) phase constant of the resultant wave? (c) Plot the wave form of the resultant wave at t=0, and discuss its behavior as tincreases.

Short answer

a) The amplitude of the resultant wave is $0.83y_1$.

b) The phase constant of the resultant wave is $37°$.

c) The wave form of the resultant wave at t = 0 is plotted, and its behavior as t increases is discussed.

Step-by-step solution

The given data

i) The wave equation is,$y(x,t)=y_{m}sin(kx-\omega t+\varphi )$

ii) Three waves have the same frequency, $f$.

iii) The amplitudes of three waves,$y_1,y_1/2,y_1/3$

iv) Phase constants of three waves, $0,\pi /2,\pi$

Understanding the concept of the wave equation

We can find the amplitude by using the phasor and taking the resultant of the given amplitudes. The phase constant of the resultant wave can be calculated from the x and y component of the resultant amplitude. The plot can be drawn by forming the wave equation for the resultant wave.

a) Calculation of the amplitude

Phasor diagram for the given waves:

Let y be the resultant of the three given waves.

The horizontal component of y is,

$$\begin{gathered} y_x=y_1-\frac{y_1}{3} \\ =\frac{2}{3}{y}_1 \end{gathered}$$

The vertical component of y is,

$y_y=\frac{y_1}{2}$

Therefore amplitude of the resultant wave is given as:

$$\begin{gathered} y_m=\sqrt{y_x^2+y_y^2} \\ =\sqrt{{\left(\frac{2}{3}{y}_1\right)}^2+{\left(\frac{y_1}{2}\right)}^2} \\ =\sqrt{\frac{4}{9}{y}_1^2+\frac{1}{4}{y}_1^2} \\ =0.83y_1 \end{gathered}$$

Therefore, the value of the amplitude of the resultant wave is $0.83y_1$.

b) Calculation of the phase constant

The phase constant of the resultant wave is given as:

$\varphi ={\tan }^{-1}\left(\frac{y_y}{y_x}\right)$

Substitute the values in the above expression, and we get

$$\begin{gathered} \varphi ={\tan }^{-1}\left(\frac{{\displaystyle \frac{y_1}{2}}}{{\displaystyle \frac{2}{3}}{y}_1}\right) \\ \varphi =\left(\frac{3}{4}\right) \\ =36.87 \\ ~37° \end{gathered}$$

Therefore, the phase constant of the resultant wave is $37°$.

c) Plotting the waveform of the resultant wave

The general equation of the wave is,

$y=y_m\sin \left(kx-\omega t+\varphi \right)$

Therefore, the equation of the resultant wave is,

localid="1660985063034" $y=0.83y_{1}sin(kx-\omega t+{37}^{°})$

The plot of the wavelocalid="1660985065802" $y=0.83y_{1}sin(kx-\omega t+{37}^{°})$at t=0

Hence, as t increases, the wave is moving in the positive x direction with wave number k and angular velocity $\omega$.