Question

Two sinusoidal waves of the same wavelength travel in the same direction along a stretched string. For wave 1,${\mathbf{y}}_{\mathbf{m}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mm}$ and$\mathbf{\varphi }\mathbf{=}\mathbf{0}\mathbf{°}$; for wave 2,${\mathbf{y}}_{\mathbf{m}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mm}$and$\mathbf{\varphi }\mathbf{=}\mathbf{70}\mathbf{°}$. What are the (a) amplitude and (b) phase constant of the resultant wave?

Short answer

a) Amplitude of the resultant wave is 6.7mm

b) Phase constant for the resultant wave is$45°$$45°$

Step-by-step solution

Given data

For wave 1$y_m=3.0mm$ and$\varphi =0°$

For wave 2 $y_m=5.0mm$and$\varphi =70°$

Understanding the concept of resonant frequency 

Here, we can find the x and y components of both waves. Using the vector addition rule, we can find the x and y components of the resultant. Using the Pythagoras theorem, we can find the magnitude.

To find the phase difference, we can use the x and y components of the resultant.

Formula:

The formula for hypotenuse using the Pythagoras theorem,$y=\sqrt{y_x^2+y_y^2}................\left(1\right)$

The phase angle can be calculated using equation, $\theta =\frac{y_y}{y_x}...........\left(2\right)$

Step 3(a): Calculation of amplitude

x component of wave 1 is given as:

$y_{x1}=3.00mm$

x component of wave 2 is given as:

$$\begin{gathered} y_{x2}=5.00\cos 70° \\ =1.71mm \end{gathered}$$

y component of wave 1 is given as:

$y_{y1}=0.00mm$

y component of wave 2 is given as:

$$\begin{gathered} y_{y2}=5.00\sin 70° \\ =4.7mm \end{gathered}$$

So x component of resultant wave is given as:

role="math" localid="1660991009082" $$\begin{gathered} y_y=y_{x1}+y_{x2} \\ =3.00+1.71 \\ =4.71mm \end{gathered}$$

So y component of resultant wave is given as:

$$\begin{gathered} y_y=y_{y1}+y_{y2} \\ =0.00+4.7 \\ =4.7mm \end{gathered}$$

So, using equation (1), the amplitude of the resultant wave is given as:

$$\begin{gathered} y=\sqrt{4.{71}^2+4.7^2} \\ =6.65mm \\ \approx 6.7mm \end{gathered}$$

Hence, the amplitude of the resultant wave is$6.7mm$

Step 4(b): Calculation of phase difference

So, the phase difference using equation (ii) is given as:

$$\begin{gathered} \tan \varphi =\frac{4.7}{4.71} \\ \varphi =45° \end{gathered}$$

Hence, the value of phase difference is$45°$