# Video: Relating the Energy of a Wave to its Amplitude

The energy of a ripple on a pond is proportional to the amplitude squared. If the amplitude of the ripple is 0.100 cm at a distance from the source of 6.0 m, what was the amplitude at a distance of 2.0 m from the source?

03:10

### Video Transcript

The energy of a ripple on a pond is proportional to the amplitude squared. If the amplitude of the ripple is 0.100 centimeters at a distance from the source of 6.0 meters, what was the amplitude at a distance of 2.0 meters from the source?

The amplitude of 0.100 centimeters weβll call π΄ sub one, and the distance from the source of 6.0 meters weβll call π sub one. The closer distance of 2.0 meters weβll call π sub two. And overall, the amplitude we want to solve for weβll call π΄ sub two.

Weβre told in the problem statement that the energy of a ripple on the pond, weβll call it πΈ, is proportional to the wave amplitude squared. We also know that as the wave moves away from the source, wave energy is proportional to one over π where π is the distance from the wave source. If we string these two proportionalities together, we can write that energy is proportional to the amplitude squared which is itself proportional to one over the distance from the wave source.

If we apply this relationship to our scenario, we see that wave amplitude squared is proportional to one over π where π is the distance from the wave source. This means that if we take the ratio of π΄ sub one squared divided by π΄ sub two squared, then this fraction is equal to one over π sub one divided by one over π sub two. The right side of this equation simplifies to π sub two divided by π sub one. We want to solve for π΄ sub two, so letβs cross multiply to get π΄ sub two by itself.

When we do that, we find π΄ sub two squared equals π΄ sub one squared times π sub one divided by π sub two. If we take the square root of both sides, this cancels out the square root and square terms on the left side, and weβre left with an equation for π΄ two that reads π΄ two equals π΄ one times the square root of π sub one divided by π sub two.

Each value in the right side of this equation is known, so letβs plug them in now. In units of meters, π΄ sub one is 0.001 meters; π sub one is 6.0 meters; and π sub two is 2.0 meters. When we calculate this value for π΄ sub two, we find that it equals 0.17 meters. Thatβs the amplitude of the wave a distance of 2.0 meters from the source.