Question

Consider two identical and symmetrical wave pulses on a string. Suppose the first pulse reaches the fixed end of the string and is reflected back and then meets the second pulse. When the two pulses overlap exactly, the superposition principle predicts that the amplitude of the resultant pulses, at that moment, will be what factor times the amplitude of one of the original pulses?a. 0b. 1c. -2d. -1

Answer 1

Answer:

The amplitude of the resultant wave will be 0.

Explanation:

Suppose the first wave has an amplitude of A. Its angle is given as wt.

The second way will also have the same amplitude as that of first.

After the reflection, a phase shift of π is added So the wave is given as

$W_1=W_2=Acos(\omega t)\\W_1^{'}=Acos(\omega t+ \pi)$

Adding the two waves give

                               $W_1'+W_2=Acos(\omega t+ \pi)+Acos(\omega t)\\W_1'+W_2=-Acos(\omega t)+Acos(\omega t)\\W_1'+W_2=0$

So the amplitude of the resultant wave will be 0.