Question

Two sinusoidal waves of the same period, with amplitudes of 5.0and 7.0 mm, travel in the same direction along a stretched string; they produce a resultant wave with an amplitude of 9.0 mm. The phase constant of the 5.0 mmwave is 0.What is the phase constant of the 7.0 mmwave?

Short answer

Phase constant of the 7 mm wave is$84°$

Step-by-step solution

The given data

i) Amplitudes of two waves, $y_{m1}=5.0\mathrm{mm}\mathrm{and}{y}_{m2}=7.0\mathrm{mm}$

ii) Amplitude of resultant wave, $y_m=9.0\mathrm{mm}$

iii) Phase constant of 5.00 mm wave, $\theta =0°$

Understanding the concept of phase

The phase constant of sinusoidal waves of the same period which travel in the same direction is given by the cosine formula, and it can be shown in a phase diagram.

Formula:

Phase constant using Cosine Formula,

$$\begin{gathered} y_m^2=y_{m^1}^2+y_{m^2}^2-2y_{m1}{y}_{m2}\mathrm{cos\theta } \\ =y_{m^1}^2+y_{m^2}^2-2y_{m1}{y}_{m2}\cos \phi \end{gathered}$$ (i)

Calculation of phase of 7.00 mm wave

Phasor diagram can be drawn as below:

As we need to find the phase constant i.e. $\mathrm{\phi }$Hence, using equation (i) and given values, we get the value of phase as:

$$\begin{gathered} \cos \phi =\frac{y_m^2-y_{m1}^2+y_{m2}^2}{2y_{m1}{y}_{m2}} \\ \cos \phi =\frac{{\left(9.0\right)}^2-{\left(5.0\right)}^2+{\left(7.0\right)}^2}{2\left(5.0\right)\left(7.0\right)} \\ \cos \phi =0.10 \\ \phi =84° \end{gathered}$$

Hence, the value of phase is$84°$