Question

Interference and Diffraction Two waves with identical frequency but different amplitudes $A_{1}=5.0 \mathrm{cm}\( and \)A_{2}=3.0 \mathrm{cm},$ occupy the same region of space (are superimposed). (a) At what phase difference does the resulting wave have the largest amplitude? What is the amplitude of the resulting wave in that case? (b) At what phase difference does the resulting wave have the smallest amplitude and what is its amplitude? (c) What is the ratio of the largest and smallest amplitudes?

Short answer

Answer: The largest amplitude occurs at a phase difference of 0 and the amplitude is 8 cm. The smallest amplitude occurs at a phase difference of π and the amplitude is 2 cm. The ratio of the largest to the smallest amplitude is 4.

Step-by-step solution

Applying the Principle of Superposition

When two waves with identical frequency are superimposed, the principle of superposition states that the resultant amplitude of the wave is the vector sum of the amplitudes of the individual waves. We can define this as: \(A = \sqrt{(A_1 \cos(\phi_1) + A_2 \cos(\phi_2))^2 + (A_1 \sin(\phi_1) + A_2 \sin(\phi_2))^2}\) where \(A\) is the amplitude of the resultant wave, \(A_1\) and \(A_2\) are the amplitudes of the individual waves, and \(\phi_1\) and \(\phi_2\) are the phase angles.

Finding the Maximum Amplitude

In order to find the maximum amplitude, we need to ensure that the amplitudes of both waves have the same phase, i.e., the phase difference \(\Delta\phi = \phi_1 - \phi_2 = 0\). So, applying the formula from the previous step, we get: \(A_\text{max} = \sqrt{(5 \cos(0) + 3 \cos(0))^2 + (5 \sin(0) + 3 \sin(0))^2} = 8\,\text{cm}\) The maximum amplitude occurs when the phase difference is \(0\), and the amplitude in that case is \(8\,\text{cm}\).

Finding the Minimum Amplitude

To find the minimum amplitude, we need to ensure that the amplitudes of both waves have opposite phase, i.e., the phase difference \(\Delta\phi = \phi_1 - \phi_2 = \pi\). So, applying the formula from step 1, we get: \(A_\text{min} = \sqrt{(5 \cos(0) - 3 \cos(0))^2 + (5 \sin(0) - 3 \sin(0))^2} = 2\,\text{cm}\) The minimum amplitude occurs when the phase difference is \(\pi\), and the amplitude in that case is \(2\,\text{cm}\).

Determining the Ratio of the Largest and Smallest Amplitudes

Now that we have the maximum and minimum amplitudes, we can find the ratio: \(r = \frac{A_\text{max}}{A_\text{min}} = \frac{8\,\text{cm}}{2\,\text{cm}} = 4\) Hence, the ratio of the largest amplitude to the smallest amplitude is 4. To sum up: (a) The largest amplitude occurs at a phase difference of \(0\) and the amplitude is \(8\,\text{cm}\). (b) The smallest amplitude occurs at a phase difference of \(\pi\) and the amplitude is \(2\,\text{cm}\). (c) The ratio of the largest to the smallest amplitude is 4.