Question 3 of 7: Sound propagation over water and a piston-type calibrator

Question 3: Sound propagation over water and a piston-type calibrator (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Sound travels across a 500 m wide lake in air at $22\ {}^\circ\text{C}$ with a listener at mid-water, and a 50 Hz piston calibrator generates 94 dB and 104 dB rms at its open end where the characteristic impedance is 407 rayl.

Find. The three travel times across the lake, and for the calibrator the motor speed, the peak pressures, the pressure–time expressions and the peak piston velocities.

Approach. Part A needs only the adiabatic speed of sound and path lengths measured off the geometry; Part B inverts the decibel definition to get rms pressure, scales by $\sqrt{2}$ for the peak of a sinusoid, and uses the plane-wave relation $u=p/\rho_0 c$ to convert pressure to piston velocity.

Part A — Echo Lake

1. Compute the speed of sound in the ambient air. Sound propagates adiabatically, so$$c=\sqrt{\gamma R T}=\sqrt{1.4\times 287\times 295.2}=\sqrt{118\,611}$$giving$$\boxed{c=344.4\ \text{m/s}}$$The molar route offered in the question, $c=\sqrt{\gamma R_u T/M}$ with $R_u=8.3145$ J/(mol·K) and $M=0.029$ kg/mol, gives 344.2 m/s; the 0.05 % difference is only because $R_u/M = 286.7$ rather than the rounded 287 J/(kg·K) quoted, and either value is acceptable.
2. Time for the call to reach the kayak. The friend sits at mid-water, $W/2 = 250$ m away, so$$t_1=\frac{W/2}{c}=\frac{250}{344.4}$$$$\boxed{t_1=0.726\ \text{s}}$$
3. Time for the caller to hear the echo. The echo path runs the full width to the far shore and back to the caller, a round trip of $2W = 1\,000$ m:$$t_2=\frac{2W}{c}=\frac{1000}{344.4}$$$$\boxed{t_2=2.904\ \text{s}}$$The delay is comfortably above the roughly 0.1 s separation the ear needs to hear an echo as a distinct event rather than as reverberation, which is why the lake earns its name.
4. Interval between the call and the echo as heard from the kayak. The echo reaches the kayak after travelling $W$ to the far shore and then $W/2$ back, i.e. 750 m, so it arrives at $t = 750/344.4 = 2.178$ s. Subtracting the arrival of the direct sound,$$\Delta t = \frac{W+W/2}{c}-\frac{W/2}{c}=\frac{W}{c}=\frac{500}{344.4}$$$$\boxed{\Delta t = 1.452\ \text{s}}$$The result is exactly one lake-width of travel, and it is independent of where on the line between the shores the friend happens to be — a neat check on the arithmetic.

Part B — Piston-type calibrator

1. Relate motor speed to tone frequency. A crank-driven piston completes exactly one pressure cycle per shaft revolution, so the required shaft speed is$$N = 60 f = 60\times 50$$$$\boxed{N=3\,000\ \text{rev/min}}$$
2. Invert the decibel definition to get rms pressures. From $L_p=20\log_{10}(p_{\text{rms}}/p_{\text{ref}})$ with $p_{\text{ref}}=20\times10^{-6}$ Pa:$$p_{\text{rms}}=p_{\text{ref}}\,10^{L_p/20}$$so $p_{\text{rms}}=20\times10^{-6}\times10^{94/20}=1.002$ Pa at 94 dB and $20\times10^{-6}\times10^{104/20}=3.170$ Pa at 104 dB. The 94 dB value being almost exactly 1 Pa is not a coincidence: 94 dB is the standard pistonphone calibration level precisely because it corresponds to 1 Pa.
3. Convert rms to peak amplitude. For a pure sinusoid the peak exceeds the rms by $\sqrt{2}$:$$\hat{p}=\sqrt{2}\,p_{\text{rms}}$$$$\boxed{\hat{p}_{94}=1.42\ \text{Pa}\qquad \hat{p}_{104}=4.48\ \text{Pa}}$$The 10 dB step between the two calibrations corresponds to a pressure ratio of $\sqrt{10}=3.16$, which the two peak values reproduce exactly.
4. Write the pressure–time expressions at the open end. With $\omega = 2\pi f = 2\pi(50)=314.2$ rad/s and taking the phase reference at $t=0$:$$p_{94}(t)=1.42\sin(314.2\,t)\ \text{Pa}$$$$p_{104}(t)=4.48\sin(314.2\,t)\ \text{Pa}$$Because the cylinder is short compared with the 6.9 m wavelength at 50 Hz, the pressure is sensibly uniform inside it and no propagation term $kx$ is needed; the cavity behaves as a lumped compliance driven by the piston.
5. Convert pressure to peak piston velocity. At the open end the radiated field is locally a plane wave, so pressure and particle velocity are in phase and related by the characteristic impedance:$$\hat{u}=\frac{\hat{p}}{\rho_0 c}$$Substituting $\rho_0 c=407$ rayl gives $\hat{u}_{94}=1.42/407$ and $\hat{u}_{104}=4.48/407$, that is$$\boxed{\hat{u}_{94}=3.48\ \text{mm/s}\qquad \hat{u}_{104}=11.0\ \text{mm/s}}$$These are very small velocities — a few millimetres per second — which is why a pistonphone can be built with a tiny eccentric drive and still deliver a level of 104 dB.