Energy and Power of a Wave Traveling Along a String — Question 6

Question 6

A finite train has energy set by its duration

A source emits $N=40$ cycles at $f=20\,\mathrm{Hz}$ with mean carried power $.30\,\mathrm{W}$ into a string with speed $v=10\,\mathrm{m/s}$. Neglect short turn-on and turn-off contributions.

1. Find the train duration, spatial length and total energy.

2. If frequency doubles while $N,A,\mu,v$ stay fixed, how do duration, train length and energy change?

Question 6 - Solution

A finite train has energy set by its duration

Method and mathematical working

Convert cycle count to emission duration. The same duration determines both train length and total energy at the stated average power.$$\begin{aligned}\Delta t&=\frac Nf,\qquad \ell=v\frac Nf,\\E&=\overline P\frac Nf.\\[4pt]\Delta t&=40/20=2\,\mathrm s,\quad\ell=10(2)=20\,\mathrm m,\quad E=0.30(2)=0.60\,\mathrm J,\\f\to2f&:\quad E\to(4\overline P)(\Delta t/2)=2E.\end{aligned}$$

(a) Calculation and conclusion

Duration is $N/f=2.0\,\mathrm{s}$, length is $vN/f=20\,\mathrm{m}$, and energy is $\overline P N/f=.60\,\mathrm{J}$. The long train assumption makes endpoint corrections negligible.

(b) Interpretation and limits

At fixed amplitude, power increases fourfold with frequency, but duration and length halve. Total energy therefore doubles, to $1.20\,\mathrm{J}$. Counting a fixed number of cycles differs from maintaining a fixed emission duration.

Physical check

For fixed $N$ and amplitude, doubling frequency quadruples power but halves duration. Total energy therefore doubles, rather than quadrupling.