Chapter 17, Problem 1Q

To determine

To find:

a) The rank of phase difference according to the average rate of transport of energy by the waves, greatest first.

b) The average rate of energy transport for the first choice in (a).

Answer to Problem 1Q

Solution:

a) The rank of phase difference according to the average rate of transport of energy by the waves, greatest first, is  $\varphi =0 , \varphi =0.2 \lambda and \varphi =0.5 \lambda$

b) The average rate of energy transport for (a) $P_{avg}=4P_{avg1}$

Explanation of Solution

1) Concept:

The rate of energy transported by a travelling wave depends on the intensity of the wave as well as the area to which the energy is transported. The intensity of a resultant wave depends on the phase difference between the two superposing waves.

2) Formula:

$P= I.A$

3) Given:

i) The average rate of energy transported by a single wave = $P_{avg1}$

ii)  The phase difference between the two waves sent through the pipe are  $\varphi =0 , \varphi =0.2 \lambda and \varphi =0.5 \lambda$

4) Calculations:

a) The rate of energy transported is given by

$P= I.A$

$I$ is intensity of the wave and $A$ is area to which the energy is transported.

When two waves are sent with phase difference ${\varphi }_1=0$ ; the two waves will coincide exactly with each other. There will be constructive interference between the two and the resultant amplitude will be twice the original amplitude. Hence, the intensity of the resultant wave will be 4 times the original. (Since $I={\left(A\right)}^2$)

Thus, the rate of energy transport for ${\varphi }_1=0$ will be more than that for a single wave.

When two waves are sent with phase difference ${\varphi }_2=0.2 \lambda$ ; the two waves will coincide partially. There will be partial constructive interference between the two waves.

Hence, the resultant intensity will be more than the single wave but less than that for ${\varphi }_1=0$

When two waves are sent with phase difference ${\varphi }_3=0.5 \lambda$ ; the two waves will show destructive interference. i.e the resultant wave will be a standing wave and not a travelling wave. Hence, it will not transport the energy.

Hence, the ranking of the situations will be

$P_{\phi =0}>P_{\phi =0.2\lambda }>P_{\phi =0.5 \lambda }$

b) Since the amplitude of the resultant wave is twice the single wave, intensity is four fold. Hence, the rate of energy transported will be

$P_{avg}=I .A$

$\therefore P_{avg}=\left( 4I_1\right).A$

$\therefore P_{avg}=4 . P_{avg 1}$

Conclusion:

The energy transported by a wave can be calculated by using the intensity of the wave. Here, the intensity changes according to the phase difference between the two superposing waves. Hence the rate of energy transported changes as the phase difference changes.