# Ultrasonic pulse in tube produces two reflections

I have an interesting problem which after much research I can not find any answers.

I have an ultrasonic transducer sending a $$500 \,\mu\text{S}$$ $$48 \,\text{kHz}$$ pulse into a clean $$20 \,\text{mm}$$ diameter PVC pipe with $$1 \,\text{mm}$$ walls.

The pipe is closed off by a plunger which can be moved up/down the pipe. The transducer receives the echo at the expected time, which matches the transmitted pulse time length. Also received a short time later is a second slightly elongated pulse about $$10 \%$$ louder than the first pulse. The second pulse delay time is not long enough to be a secondary reflection.

I also know the second pulse is not a wave travelling down the pipe walls because cotton wool in the pipe completely kills both pulses.

Typical times of pulses are:

1. Plunger at $$200 \,\text{cm}$$, pulse $$A$$ time from Tx pulse $$11.6 \,\text{mS}$$, pulse $$B$$ $$12.8 \,\text{mS}$$.
2. Plunger at $$250 \,\text{cm}$$, pulse $$A$$ time from Tx pulse $$14.5 \,\text{mS}$$, pulse $$B$$ $$16.1 \,\text{mS}$$.
3. Plunger at $$200 \,\text{cm}$$, pulse $$A$$ time from Tx pulse $$17.2 \,\text{mS}$$, pulse $$B$$ $$19.2 \,\text{mS}$$.

Any brainstorming thoughts?

Below is a measurement at $$3$$ meters:

I have added an ultra-sonic microphone to the end of the $$3$$ meter pipe and removed the plunger. You can see the two pulses but they are only $$1 \,\text{mS}$$ apart. The reflected pulses at the transducer are $$2 \,\text{mS}$$ apart, as shown above. This suggests the two pulses are travelling at different speeds??? which also agrees with the results table above. NOTE: Holding the pipe with your hands does not affect the pulse shape which seems to also suggest the pulses are not travelling down the pipe walls.

I guess there is also the possibility that the second pulse is travelling in a different mode which has a longer distance to travel such as bouncing off the internal walls.

• In the experiment with the cotton wool, where is the cotton wool placed? inside or outside? near end, or far end, or through out the pipe?
– AJN
Commented Apr 22 at 12:27
• Try changing things to see what affects the second pulse. What happens if you get a longer PCV pipe? What happens if you use a different plunger? Commented Apr 22 at 14:07
• The cotton wool was place at the transducer end. Care was taken to not compact the cotton wool. Commented Apr 22 at 14:52
• I think I'd try damping mechanical coupling by wrapping the pipe with some soft material. Commented Apr 22 at 15:09
• Maybe the transducer is actually sending those two pulses! There could be some electronic problem with the pulse shape. Or maybe the pulse is travelling through the plunjer and then reflected from the back side of it, adding to the front side reflection. Commented Apr 22 at 17:14

Let us assume the walls are rigid and have an inner radius of $$a$$. Then the lowest modes of propagation in the air would be the plane wave mode and the first axi-symmetric mode. Assuming axi-symmetry, we may write a complex pressure mode as $$\phi_n(r,z) = J_0(k_{rn}r)e^{ik_{zn}z}$$ where we have assumed an $$e^{-i\omega t}$$ time dependence, $$\omega$$ is the angular frequency, $$t$$ is the time, $$r$$ is the radial position, $$z$$ is the position along the tube, $$\phi_n$$ is the mode function, and where $$k_{rn}$$ is a radial wavenumber and $$k_{zn}$$ is the on-axis wavenumber. The mode and wavenumbers have been indexed with an $$n$$ because the modes are discrete. Since we are assuming the walls are rigid, we know that the radial gradient at $$r=a$$ must go to zero, leading to the requirement $$-k_{rn}J_1(k_{rn}a) = 0,$$ or $$k_{rn} = \frac{j_{1,n}}{a}$$ where $$j_{m,n}$$ is the $$n$$th zero of the $$m$$th order Bessel J function.
The wavenumbers are related by the equation $$k_0^2 = \frac{\omega^2}{c_0^2} = k_{rn}^2 + k_{zn}^2,$$ where $$c_0$$ is the speed of sound in air, and so we may write $$k_{zn} = \pm \sqrt{\frac{\omega^2}{c_0^2} - \frac{j_{1,n}^2}{a^2}}.$$ Thus, the phase speed of the $$n$$th mode may be written as $$c_{\text{ph},n} \equiv \frac{\omega}{k_{zn}} = \frac{c_0}{\sqrt{1 - \frac{j_{1,n}^2c_0^2}{a^2\omega^2}}}.$$ For high frequencies the phase speed for the first few modes are increasing, which at first blush looks like they will not be able to explain the delayed pulse. However, pulses propagate at the group speed, not the phase speed. The group speed is defined using derivatives: $$c_{\text{gr},n} \equiv \left(\frac{\partial k_{zn}}{\partial\omega}\right)^{-1} = c_0\sqrt{1 - \frac{j_{1,n}^2c_0^2}{a^2\omega^2}},$$ which decreases with increasing $$n$$ for the first few $$n$$.
The value of $$j_{1,1}$$ is about 3.83, $$j_{1,2}$$ is about 7.01, $$a$$ is 10 mm, $$c_0$$ is about 343 m/s, and $$\omega$$ is about 301593 rad/s. Then we would expect to see $$c_{\text{gr},0} = c_0 = 343$$ m/s, $$c_{\text{gr},1} \approx 309$$ m/s, and $$c_{\text{gr},2} \approx 207$$ m/s. Using the 200 cm plunger, which leads to a 400 cm travel time, we can then expect arrivals at 11.66 ms, 12.96 ms, and 19.33 ms for these three modes, respectively. The reported arrival times for the first two pulses are 11.6 ms and 12.8 ms.