Question

1. shown are four different pairs of transverse wave pulses that move toward each other. at some point in time, the pulses meet and interact (interfere) with each other. rank the four cases, from most to least, on the basis of the height of the peak that results when the centers of the pulses coincide.

Explanation:

Step1: Analyze case A

When the centers of the pulses in case A coincide, it is constructive interference. The height of the peak is the sum of the heights of the two individual pulses. Let the height of the smaller pulse be \(h_1\) and the larger one be \(h_2\). The resultant height \(H_A=h_1 + h_2\) (where \(h_2>h_1\))

Step2: Analyze case B

For case B, when the centers of the pulses coincide, it is also constructive interference. The two pulses have the same height \(h\). Using the principle of superposition, the resultant height \(H_B = h+h=2h\)

Step3: Analyze case C

In case C, when the centers of the pulses coincide, it is destructive interference. The amplitudes of the two pulses are equal in magnitude but opposite in sign. So \(H_C=0\)

Step4: Analyze case D

For case D, when the centers of the pulses coincide, it is destructive interference. Let the magnitude of the amplitude of the positive - going pulse be \(h\) and the negative - going pulse be \(H\) (\(H > h\)). The resultant height \(H_D=H - h\) (negative value, but we consider the magnitude of the peak. Since \(H - h>0\) as \(H>h\))

Since \(H_A> H_B>H_D > H_C\) (assuming appropriate relative amplitudes from the figure's visual cues where case A has a large - small combination, case B has equal - height constructive, case D has unequal - height destructive and case C has equal - height destructive)

Answer:

A, B, D, C