Energy and Power of a Wave Traveling Along a String — Question 5

Question 5

Equal displacement amplitude does not imply equal carried power

Two strings have the same tension, but linear densities $\mu$ and $9\mu$. Separate traveling waves have the same frequency.

1. For equal displacement amplitudes, compare their mean powers.

2. Find the amplitude ratio required for equal powers, and explain why the slower string can carry more power at equal amplitude.

Question 5 - Solution

Equal displacement amplitude does not imply equal carried power

Method and mathematical working

Combine density and speed into the string impedance. Equal tension does not imply equal power for equal displacement amplitudes.$$\begin{aligned}\mu v&=\sqrt{F\mu},\qquad \overline P=\tfrac12\sqrt{F\mu}\,\omega^2A^2,\\\frac{\overline P_2}{\overline P_1}&=3\frac{A_2^2}{A_1^2}.\\[4pt]\overline P_2=\overline P_1&\Rightarrow 3A_2^2=A_1^2\Rightarrow A_2/A_1=1/\sqrt3.\end{aligned}$$

(a) Calculation and conclusion

$\overline P=\tfrac12\sqrt{F\mu}\,\omega^2A^2$. The denser string has three times the factor $\sqrt{F\mu}$, hence three times the power at equal $A,\omega$, although its wave speed is one third as large.

(b) Interpretation and limits

Equal powers require $3A_2^2=A_1^2$, so $A_2/A_1=1/\sqrt3$. The denser string contains nine times as much oscillating mass per unit length, more than compensating its threefold smaller transport speed.

Physical check

The denser string moves waves more slowly but has a larger mechanical impedance. At equal power its amplitude must decrease, consistent with the fourth-root density dependence.