The Doppler Effect — Question 9

Question 9

Accelerating pulse sources require the displacement between emissions

A source moves toward a fixed observer at $x=D=100\,\mathrm m$ with $x_s(t)=10t+10t^2$ in SI units. It emits two pulses at $t=0$ and $t=0.100\,\mathrm s$. Sound propagates rightward at $c=340\,\mathrm{m/s}$.

1. Find the arrival-time interval and its reciprocal.

2. Compare with inserting the initial source speed into a steady-motion Doppler formula. Which velocity exactly describes this pair of pulses?

Question 9 - Solution

Accelerating pulse sources require the displacement between emissions

(a) Method and calculation

The source moves $\Delta x_s=10(0.1)+10(0.1)^2=1.10\,\mathrm m$. Arrival times satisfy $t_r=t_e+(D-x_s)/c$, so$$\Delta t_r=0.100-\frac{1.10}{340}=0.0967647\,\mathrm s,$$$$1/\Delta t_r=10.33435\,\mathrm{s^{-1}}.$$

(b) Interpretation and further calculation

The initial speed is $10\,\mathrm{m/s}$, which would predict $10(340)/(340-10)=10.30303\,\mathrm{s^{-1}}$. The exact pair calculation uses the average source velocity$$\overline v=\Delta x_s/\Delta t_e=11.0\,\mathrm{m/s},\qquad\Delta t_r=\Delta t_e(1-\overline v/c).$$The reciprocal of this finite arrival interval is a pair rate, not a claim of constant received frequency.

Physical check

The observer distance cancels from the interval because both fronts propagate through the same uniform medium along the same direction.