Question

Two waves with identical frequency but different amplitudes $A_{1}=6.0 \mathrm{cm}\( and \)A_{2}=3.0 \mathrm{cm},$ occupy the same region of space (i.e., are superimposed). (a) At what phase difference will the resulting wave have the highest intensity? What is the amplitude of the resulting wave in that case? (b) At what phase difference will the resulting wave have the lowest intensity and what will its amplitude be? (c) What is the ratio of the two intensities?

Short answer

Answer: The phase difference values for maximum intensity are \(0, 2\pi, 4\pi, \dots\) and for minimum intensity are \(\pi, 3\pi, 5\pi, \dots\). The amplitude of the resulting wave at maximum intensity is 9 cm and at minimum intensity is 3 cm. The ratio of the maximum to minimum intensities is 9.

Step-by-step solution

Use the principle of superposition

By the principle of superposition, the resulting wave's displacement y is the vector sum of the displacements of the two individual waves: \(y = y_1 + y_2 = A_1 \sin{(kx - \omega t)} + A_2 \sin{(kx - \omega t + \phi)}\)

Find phase difference for maximum intensity

For maximum intensity, the amplitude of the resulting wave should be maximum. It occurs when the two waves are in phase (\(\phi = 0, 2\pi, 4\pi, \dots)\). When the two waves add up constructively, the resulting amplitude is the sum of their amplitudes: \(A_{max} = A_1 + A_2 = 6 \,\text{cm} + 3\,\text{cm} = 9\, \text{cm}\)

Find phase difference for minimum intensity

For minimum intensity, the amplitude of the resulting wave should be minimum. It occurs when the two waves are out of phase (\(\phi = \pi, \3\pi, 5\pi, \dots)\). When the two waves have destructive interference, the resulting amplitude is the difference of their amplitudes: \(A_{min} = |A_1 - A_2| = |6\, \text{cm} - 3\, \text{cm}| = 3\, \text{cm}\)

Calculate intensities

Intensity is proportional to the square of the amplitude. To find the ratio of the intensities, calculate the intensities for maximum and minimum amplitude cases, and divide one by the other: \(I_{max} \propto A_{max}^2 = (9\, \text{cm})^2 = 81\,\text{cm}^2\) \(I_{min} \propto A_{min}^2 = (3\, \text{cm})^2 = 9\,\text{cm}^2\)

Calculate ratio of intensities

To find the ratio of the maximum and minimum intensities, divide \(I_{max}\) by \(I_{min}\): \(I_{ratio} = \frac{I_{max}}{I_{min}} = \frac{81\,\text{cm}^2}{9\, \text{cm}^2} = 9\) The resulting wave has the highest intensity when the phase difference is \(0, 2\pi, 4\pi, \dots\), and its amplitude is \(9\,\text{cm}\). It has the lowest intensity when the phase difference is \(\pi, 3\pi, 5\pi, \dots\), and its amplitude is \(3\,\text{cm}\). The ratio of the maximum to minimum intensities is \(9\).