91Ó°ÊÓ

For wave 1$y_m=3.0mm$ and$\mathrm{ϕ}=0°$

For wave 2 $y_m=5.0mm$and$\mathrm{ϕ}=70°$

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, $\mathrm{θ}=\frac{y_y}{y_x}...........\left(2\right)$

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:

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

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

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

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

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