How to Calculate Echo Distance from a Return Time

A microphone records a clap at \(0.120\,\text{s}\) and its echo at \(0.320\,\text{s}\). If sound travels at \(340\,\text{m/s}\), is the reflecting wall 34 m or 68 m away?

The wall is 34 m away. The sound has travelled 68 m, but that distance includes the journey to the wall and back.

To calculate echo distance from a return time, use:

\[
d=\frac{v\Delta t}{2}
\]

where:

• \(d\) is the one-way distance to the reflecting surface, in metres (m).
• \(v\) is the speed of sound, in metres per second (m/s).
• \(\Delta t\) is the elapsed time between the original sound and its echo, in seconds (s).

For this recording:

\[
\begin{aligned}
\Delta t&=0.320-0.120\\
&=0.200\,\text{s}\\[4pt]
d&=\frac{340(0.200)}{2}\\
&=34.0\,\text{m}
\end{aligned}
\]

The division by two is essential. It converts the total distance travelled by the sound into the distance between the microphone and the wall.

This method assumes the sound travels directly to a stationary reflecting surface and returns along the same path, with the source and microphone at approximately the same position.

01Why an echo takes twice the one-way travel time

Imagine clapping in a long corridor with a hard wall at the far end.

The clap creates a brief disturbance in the surrounding air. That disturbance travels towards the wall, reflects, and returns to your position.

Before introducing the mathematics, predict what would happen if the wall were moved twice as far away.

Would the echo take twice as long, or four times as long?

It would take twice as long, assuming the speed of sound remains constant. Both the outward journey and the return journey become twice as long.

What actually travels through the air?

Sound in air is a longitudinal mechanical wave.

As a sound wave passes, air particles oscillate backwards and forwards approximately parallel to the direction in which the wave travels.

The particles do not travel all the way from the speaker to the wall and back. Instead, they oscillate around their equilibrium positions, transferring the disturbance through the air.

A useful model is a stretched slinky. Push one end forwards briefly, and a compressed region travels along the slinky. Individual coils move forwards and backwards, but they do not follow the compressed region along its entire journey.

Air behaves similarly, although air particles are not connected by springs. The disturbance is transmitted through changes in pressure and particle interactions.

A compression is a region where air pressure is temporarily higher than the surrounding equilibrium pressure. A rarefaction is a region where it is temporarily lower.

When the sound reaches a rigid wall, air cannot move through the wall. The boundary forces the travelling disturbance to reflect.

For a compression pulse hitting an ideal rigid wall, the reflected pressure pulse is also a compression.

The energy travels back towards the source through the air. That returning disturbance is the echo.

Where does the factor of two come from?

Suppose the wall is a distance \(d\) from the microphone.

The outgoing sound travels \(d\), and the reflected sound travels another \(d\).

Therefore, the total path length is:

\[
s=2d
\]

For constant speed, distance travelled equals speed multiplied by time:

\[
s=v\Delta t
\]

Combining the relationships:

\[
\begin{aligned}
2d&=v\Delta t\\
d&=\frac{v\Delta t}{2}
\end{aligned}
\]

The sound’s total distance travelled is not the same as its displacement. If the sound returns to its starting position, its net displacement is zero, even though it has travelled a substantial distance.

For the broader distinction between these quantities, see Distance and Displacement in HSC Physics Explained Clearly.

02Measuring an echo: why the first experimental design can fail

The echo formula is straightforward once the return time is known.

The harder experimental problem is identifying exactly when the reflected sound arrives.

Consider an investigation designed to measure the distance to a wall using sound.

First attempt: use a continuous sound

Place a loudspeaker and microphone near one end of a corridor, facing a rigid wall.

Connect the microphone to a device that records how air pressure changes with time.

Suppose the speaker produces a continuous tone.

The travelling sound reaches the wall, reflects, and returns to the microphone. We might predict that the recording will reveal the original sound followed by a delayed copy.

But there’s a problem.

The speaker keeps producing sound while the reflection returns. The outgoing and returning waves overlap, so the microphone measures their combined pressure variation.

Rather than two clean, separate pulses, the recording contains interference between the waves.

A continuous tone therefore does not provide a single, easily identifiable return time.

Revised design: send a short pulse

Replace the continuous tone with a short sound pulse.

The apparatus now has four essential parts:

1. A loudspeaker that produces the pulse.
2. A microphone placed very close to the loudspeaker.
3. A rigid reflecting wall some distance away.
4. A recording device connected to the microphone.

The loudspeaker sends the pulse towards the wall.

The microphone records the initial sound, followed by a weaker reflected pulse after a delay.

Because the pulse is brief, the two signals can be separated in time if the wall is sufficiently far away and the recording has adequate time resolution.

Illustrative experimental readings

The following values are invented teaching data, not measurements from a performed experiment.

The observation predicted by the reflection model is a second pulse delayed by the time needed to travel to the wall and back.

However, detecting a later pulse does not automatically prove that it came from the intended wall. Reflections from the floor, ceiling, or nearby objects might also appear.

A stronger investigation would change the wall’s position while keeping the speaker, microphone, and other conditions fixed.

For example, moving the wall 1.7 m farther away should increase the return time by:

\[
\begin{aligned}
\Delta t_{\text{extra}}
&=\frac{2(1.7)}{340}\\
&=0.010\,\text{s}
\end{aligned}
\]

The factor of two appears again because the extra 1.7 m is travelled in both directions.

If a recorded pulse moves later by the predicted amount, that supports the interpretation that it came from the wall.

This is the experimental principle behind time-of-flight measurement: identify a returning signal, measure its delay, and use the propagation speed to infer distance.

03What changes when the speaker and microphone are separated?

The familiar echo formula is simplest when the source and receiver are together.

Move the microphone away from the speaker, and the interpretation of the recorded times can change.

Consider a straight corridor with three objects in this order:

• A loudspeaker at the near end.
• A microphone farther along the corridor.
• A rigid reflecting wall beyond the microphone.

The microphone receives the direct sound first.

Later, it receives sound that travelled past the microphone, reflected from the wall, and returned.

The time between those two arrivals represents the extra journey from the microphone to the wall and back.

04What happens to the sound wave at different boundaries?

So far, we’ve assumed a hard wall.

But an echo can also occur at the open end of a tube, where sound moves from the narrow tube into a much larger space.

The returning sound behaves differently depending on the boundary.

Suppose a compression pulse travels towards the end of an air-filled tube.

Would it return as another compression if the end were open?

Not necessarily. At an ideal open end, the reflected pressure pulse is inverted: a compression reflects as a rarefaction.

To understand why, consider what the air is allowed to do at each boundary.

A rigid, closed end

At a rigid end, the air immediately next to the wall cannot move through it.

Its longitudinal displacement is therefore zero.

In a standing-wave pattern, the rigid end is a displacement node.

However, pressure can vary strongly there. The incident and reflected pressure disturbances reinforce one another at the boundary, producing a pressure antinode.

For a travelling pressure pulse, a compression reflects as a compression.

An open end

At an ideal open end, the air can move backwards and forwards as it interacts with the surrounding atmosphere.

The pressure variation at the opening is approximately zero because the pressure there remains close to atmospheric pressure.

The open end is therefore approximately a pressure node and a displacement antinode.

To satisfy the pressure condition, a reflected pressure disturbance must oppose the incident pressure disturbance at the boundary.

A compression pulse consequently reflects as a rarefaction in the ideal model.

The distinction is worth keeping clear:

Here, a node is a location with zero oscillation amplitude, while an antinode is a location with maximum oscillation amplitude for that standing-wave pattern.

The word displacement in this table refers to the tiny back-and-forth movements of air particles, not the total distance travelled by the sound wave.

How reflection leads to standing waves

An isolated sound pulse is useful for measuring echo time.

A sustained tone can reveal something different.

When an outgoing wave and a reflected wave of the same frequency travel in opposite directions, they can superpose to form a standing wave.

Instead of a pulse moving down the tube, the resulting pattern has fixed positions of nodes and antinodes.

For an ideal tube closed at one end and open at the other, the fundamental standing-wave pattern has:

• A displacement node and pressure antinode at the closed end.
• A displacement antinode and pressure node at the open end.

The distance from a displacement node to its nearest displacement antinode is one-quarter of a wavelength.

For the lowest-frequency mode, that distance is the tube length \(L\).

Therefore:

\[
L=\frac{\lambda_1}{4}
\]

where \(\lambda_1\) is the wavelength of the fundamental mode.

Since wave speed equals frequency multiplied by wavelength, \(v=f\lambda\), the fundamental frequency is:

\[
f_1=\frac{v}{4L}
\]

Here, \(f_1\) is measured in hertz (Hz), and \(L\) is the tube length in metres.

Only odd harmonics occur in the ideal closed-open tube:

\[
f_n=\frac{nv}{4L}
\]

where \(n=1,3,5,\ldots\).

For comparison, an ideal tube rigidly closed at both ends has displacement nodes at both ends and a fundamental frequency of \(v/(2L)\).

Real open ends are less ideal. Some sound radiates out, reflection is incomplete, and the effective acoustic length can extend slightly beyond the physical opening.

These effects matter in precise resonance measurements.

For basic echo-distance calculations, the central principle remains the same: measure the actual travel time and use the speed of sound along the path.

The pressure sign may change at reflection, but the elapsed travel time remains positive.

05Choosing the right calculation from an echo recording

Before substituting numbers, identify what the recorded time represents.

The calculations also rely on assumptions.

A stationary wall, a known propagation speed, identifiable pulses, and a well-defined reflection path make the usual echo formula reliable.

Temperature changes can affect the speed of sound. Wind can make outward and return travel speeds different. Multiple reflections can also produce later pulses that have travelled much farther than a simple round trip to one surface.

For the questions below, take the stated speeds and idealised reflection conditions as given. Unless stated otherwise, ignore attenuation, delays at boundaries, and additional reflections.

For one-dimensional diagrams, take the positive position direction as towards the right. Distances are reported as non-negative magnitudes. A negative pressure pulse indicates a rarefaction relative to atmospheric pressure, not a negative distance.

06Questions and solutions

Question 1

A student claps while standing beside a microphone in a large empty hall.

The microphone records the clap at \(0.600\,\text{s}\) and the first echo from a flat wall at \(0.750\,\text{s}\).

Assuming the speed of sound is \(340\,\text{m/s}\), calculate the distance from the student to the wall.

Solution 1

The wall is 25.5 m from the student.

The microphone and sound source are at the same position, so the delay is the time for the sound to travel to the wall and back.

First calculate the elapsed time:

\[
\begin{aligned}
\Delta t&=0.750-0.600\\
&=0.150\,\text{s}
\end{aligned}
\]

Then apply the echo-distance formula:

\[
\begin{aligned}
d&=\frac{v\Delta t}{2}\\
&=\frac{340(0.150)}{2}\\
&=25.5\,\text{m}
\end{aligned}
\]

The sound travels a total of 51.0 m during the round trip, but the wall is only 25.5 m away.

Multiplying the speed by the delay without dividing by two would confuse the total path length with the one-way distance.

Question 2

A loudspeaker S and microphone M are placed along a straight corridor.

The microphone lies between the loudspeaker and a rigid wall.

The speed of sound is \(340\,\text{m/s}\).

The microphone receives:

• The direct sound \(0.040\,\text{s}\) after emission.
• A reflected pulse from the wall \(0.240\,\text{s}\) after emission.

Calculate:

a) The loudspeaker-to-microphone distance.

b) The microphone-to-wall distance.

c) The loudspeaker-to-wall distance.

A student argues that part (c) should be \(340(0.240)/2=40.8\,\text{m}\). Explain the error.

Solution 2

The loudspeaker is 13.6 m from the microphone and 47.6 m from the wall. The microphone is 34.0 m from the wall.

Part (a): Loudspeaker-to-microphone distance

The direct sound travels from S to M only once.

Using \(v=340\,\text{m/s}\) and \(t=0.040\,\text{s}\):

\[
\begin{aligned}
a&=vt\\
&=340(0.040)\\
&=13.6\,\text{m}
\end{aligned}
\]

Part (b): Microphone-to-wall distance

The reflected sound travels along the path S to wall to M.

Compared with the direct S-to-M sound, it travels an additional round trip between M and the wall.

The additional travel time is:

\[
\begin{aligned}
\Delta t&=0.240-0.040\\
&=0.200\,\text{s}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
b&=\frac{v\Delta t}{2}\\
&=\frac{340(0.200)}{2}\\
&=34.0\,\text{m}
\end{aligned}
\]

Part (c): Loudspeaker-to-wall distance

Add the two distances:

\[
\begin{aligned}
D&=a+b\\
&=13.6+34.0\\
&=47.6\,\text{m}
\end{aligned}
\]

The student’s 40.8 m result is incorrect because the loudspeaker and microphone are not at the same position.

The \(0.240\,\text{s}\) interval covers the entire reflected path from S to the wall and then back to M. It is not a round trip between S and the wall.

We can verify the correct answer using that full path:

\[
\begin{aligned}
s&=47.6+34.0\\
&=81.6\,\text{m}\\[4pt]
t&=\frac{81.6}{340}\\
&=0.240\,\text{s}
\end{aligned}
\]

The calculated travel time matches the recording.

Question 3

A loudspeaker S and microphone M are separated by 6.0 m along a line parallel to a large, flat, rigid wall.

Both devices are the same unknown perpendicular distance \(h\) from the wall.

The microphone receives the direct sound first. A pulse reflected once from the wall arrives \(0.050\,\text{s}\) later.

The speed of sound is \(340\,\text{m/s}\).

Assume the sound reflects specularly from the wall, so the angle of incidence equals the angle of reflection.

Calculate the perpendicular distance \(h\).

Explain why using \(h=v\Delta t/2\) would give the wrong answer.

Solution 3

The perpendicular distance to the wall is approximately 11.1 m, not 8.5 m.

The tempting calculation \(340(0.050)/2=8.5\,\text{m}\) is wrong because the sound does not travel directly out and back along a perpendicular line from a single point.

The source and microphone are separated, and the reflected sound follows a diagonal path.

Step 1: Determine the direct path length.

The source and microphone are 6.0 m apart, so the direct sound travels 6.0 m.

The reflected sound arrives 0.050 s later.

The extra distance travelled by the reflected sound is:

\[
\begin{aligned}
\Delta s&=v\Delta t\\
&=340(0.050)\\
&=17.0\,\text{m}
\end{aligned}
\]

Step 2: Find the full reflected path length.

\[
\begin{aligned}
s_{\text{reflected}}&=6.0+17.0\\
&=23.0\,\text{m}
\end{aligned}
\]

Because S and M are the same perpendicular distance from a flat wall, the specular reflection point is halfway along the wall between their perpendicular projections.

The two reflected path segments therefore have equal lengths.

Each segment is:

\[
\frac{23.0}{2}=11.5\,\text{m}
\]

Step 3: Use the geometry.

Each segment is the hypotenuse of a right-angled triangle.

The horizontal side parallel to the wall is half the source-microphone separation:

\[
\frac{6.0}{2}=3.0\,\text{m}
\]

Using Pythagoras’ theorem:

\[
h^2+3.0^2=11.5^2
\]

\[
\begin{aligned}
h&=\sqrt{11.5^2-3.0^2}\\
&=\sqrt{123.25}\\
&\approx11.1\,\text{m}
\end{aligned}
\]

The wall is approximately 11.1 m away perpendicularly.

The result is larger than 8.5 m because the 0.050 s delay represents the difference between two unequal paths, not an entire round trip directly perpendicular to the wall.

The correct method must include both the direct 6.0 m path and the geometry of the reflected path.

Question 4

A sound pulse travels through a straight tube containing two sections of air maintained at different temperatures.

In section A, the speed of sound is \(340\,\text{m/s}\).

In section B, the speed of sound is \(320\,\text{m/s}\).

A loudspeaker and microphone are together at the entrance to section A. Section B begins at a sharp temperature boundary, and a rigid reflecting wall closes the far end of section B.

The microphone records two echoes:

Assume the temperature boundary reflects a small amount of sound and transmits the rest. Ignore multiple reflections, losses, and any extra time delay at the boundary.

Calculate:

a) The length of section A.

b) The length of section B.

c) The total distance from the loudspeaker to the rigid end wall.

A student instead uses \(340(0.054)/2\) for the total distance. Explain why this method is invalid.

Solution 4

Section A is 3.40 m long, section B is 5.44 m long, and the rigid wall is 8.84 m from the loudspeaker.

The two echoes provide separate information about how long the sound spends in each section.

Part (a): Length of section A

The first echo comes from the temperature boundary.

It travels through section A to the boundary and then returns through the same section.

Using \(v_A=340\,\text{m/s}\):

\[
\begin{aligned}
d_A&=\frac{v_A t_A}{2}\\
&=\frac{340(0.020)}{2}\\
&=3.40\,\text{m}
\end{aligned}
\]

Part (b): Length of section B

The end-wall echo spends time travelling through both sections.

Its complete return time is:

\[
0.054\,\text{s}
\]

The portion associated with section A is already known:

\[
0.020\,\text{s}
\]

Subtracting gives the time spent travelling through section B in both directions:

\[
\begin{aligned}
t_B&=0.054-0.020\\
&=0.034\,\text{s}
\end{aligned}
\]

Since \(v_B=320\,\text{m/s}\):

\[
\begin{aligned}
d_B&=\frac{v_B t_B}{2}\\
&=\frac{320(0.034)}{2}\\
&=5.44\,\text{m}
\end{aligned}
\]

Part (c): Total distance to the rigid wall

\[
\begin{aligned}
D&=d_A+d_B\\
&=3.40+5.44\\
&=8.84\,\text{m}
\end{aligned}
\]

The student’s proposed calculation gives:

\[
\frac{340(0.054)}{2}=9.18\,\text{m}
\]

That value is too large because the student has assumed sound travels at \(340\,\text{m/s}\) throughout the entire tube.

In reality, sound travels more slowly in section B.

The complete travel-time relationship is:

\[
t=\frac{2d_A}{v_A}+\frac{2d_B}{v_B}
\]

Substituting the calculated distances:

\[
\begin{aligned}
t&=\frac{2(3.40)}{340}\\
&\quad+\frac{2(5.44)}{320}\\
&=0.020+0.034\\
&=0.054\,\text{s}
\end{aligned}
\]

This confirms the recorded return time.

The central lesson is that sound speed must be matched to the region in which the sound is travelling.

A single echo delay does not always correspond to one constant propagation speed.

Question 5

An ideal narrow tube contains air in which sound travels at \(340\,\text{m/s}\).

The left end is rigidly closed. The right end is open.

A tiny pulse source and pressure microphone are placed together at an unknown position inside the tube.

The source emits a short compression pulse equally in both directions.

Assume:

• The tube is uniform, one-dimensional, and lossless.
• The rigid end reflects pressure pulses without inversion.
• The open end reflects pressure pulses with inversion.
• Both reflections have equal magnitude.
• The source and microphone do not introduce additional reflections.
• Only the first return from each end is considered when analysing the pulse recording.
• End corrections are negligible.

At the original source position, the microphone records no net first-return pressure pulse, even though both ends reflect sound.

The source and microphone are then moved exactly 0.34 m towards the rigid end.

At the new position, the microphone records:

• A positive pressure pulse 0.0040 s after emission.
• A negative pressure pulse 0.0080 s after emission.

Answer the following.

a) Determine the tube’s total length.

b) Determine the original position of the source, measured from the rigid end.

c) Predict the time at which the first reflections reach the original source position, and explain why no net pressure pulse is recorded there.

d) Calculate the fundamental standing-wave frequency and the next allowed resonant frequency.

e) A student concludes that the absence of an echo at the original position proves that the tube cannot support standing waves. Assess this conclusion.

Solution 5

The tube is 2.04 m long, and the original source was exactly at its midpoint, 1.02 m from the rigid end.

At that original position, both reflected pulses return after 0.0060 s. Their pressure variations have equal magnitudes and opposite signs, so they cancel.

The tube can still support standing waves. Its fundamental frequency is approximately 41.7 Hz, and its next allowed resonance is 125 Hz.

Part (a): Determine the tube length

Begin with the reflections recorded at the new position.

The positive pressure pulse must be the reflection from the rigid end, because a compression reflects as a compression there.

The negative pulse must be from the open end, where the returning pressure variation is inverted.

The distance to the rigid end is:

\[
\begin{aligned}
d_{\text{rigid}}
&=\frac{340(0.0040)}{2}\\
&=0.680\,\text{m}
\end{aligned}
\]

The distance to the open end is:

\[
\begin{aligned}
d_{\text{open}}
&=\frac{340(0.0080)}{2}\\
&=1.360\,\text{m}
\end{aligned}
\]

The source lies between the two ends.

Therefore, the tube length is:

\[
\begin{aligned}
L&=0.680+1.360\\
&=2.040\,\text{m}
\end{aligned}
\]

Part (b): Recover the original source position

The source was moved 0.34 m towards the rigid end.

Its original distance from the rigid end must therefore have been:

\[
\begin{aligned}
x_{\text{original}}
&=0.680+0.340\\
&=1.020\,\text{m}
\end{aligned}
\]

The distance from the original position to the open end is:

\[
\begin{aligned}
2.040-1.020
&=1.020\,\text{m}
\end{aligned}
\]

The original source was exactly halfway between the ends.

Part (c): Explain the missing echo

Both ends are 1.02 m from the original position.

The reflected pulses therefore have identical round-trip travel times.

\[
\begin{aligned}
t&=\frac{2d}{v}\\
&=\frac{2(1.020)}{340}\\
&=0.0060\,\text{s}
\end{aligned}
\]

Both first reflections arrive after 6.0 ms.

Now consider their pressure variations.

The pulse reflected from the rigid end returns as a positive pressure disturbance.

The pulse reflected from the open end returns as a negative pressure disturbance.

Under the ideal assumptions, they have equal magnitudes and arrive simultaneously.

The total pressure variation is the sum of the two disturbances:

\[
\Delta p_{\text{total}}
=\Delta p_{\text{rigid}}
+\Delta p_{\text{open}}
\]

If the positive pulse has amplitude \(+A\), the other has amplitude \(-A\).

Here, \(A\) represents the pressure-pulse amplitude in pascals (Pa).

Thus:

\[
\begin{aligned}
\Delta p_{\text{total}}
&=A+(-A)\\
&=0
\end{aligned}
\]

The microphone measures the combined pressure variation, so it records no net first-return pulse.

This is destructive interference, not an absence of reflection.

The surprising result is that two genuine echoes can arrive at exactly the expected time yet produce no net pressure signal at the microphone.

Moving the microphone away from the midpoint changes the distances to the two ends. The reflections then arrive at different times and become distinguishable.

Part (d): Calculate the resonant frequencies

The tube has one rigid end and one open end.

For the fundamental standing-wave mode, the displacement is zero at the rigid end and maximum at the open end.

The tube length therefore equals one-quarter of the fundamental wavelength:

\[
L=\frac{\lambda_1}{4}
\]

The fundamental frequency is:

\[
\begin{aligned}
f_1&=\frac{v}{4L}\\
&=\frac{340}{4(2.040)}\\
&\approx41.7\,\text{Hz}
\end{aligned}
\]

The ideal closed-open tube supports only odd harmonics.

The next allowed resonance is the third harmonic:

\[
\begin{aligned}
f_3&=3f_1\\
&=3\left(\frac{340}{8.160}\right)\\
&=125\,\text{Hz}
\end{aligned}
\]

There is no second-harmonic resonance in this ideal closed-open tube.

Part (e): Does the missing echo rule out standing waves?

No. The student’s conclusion confuses the cancellation of two particular travelling pulses with the conditions for standing-wave resonance.

The pulse experiment measures the pressure produced by short, reflected disturbances at one position and time.

At the original midpoint, the first returning pressure pulses happen to have equal and opposite amplitudes. Their sum is zero at that instant.

A standing wave involves sustained superposition of oppositely travelling waves at an allowed frequency.

Its pattern is determined by the boundary conditions at the ends of the tube.

At the rigid end, longitudinal displacement must be zero. At the open end, pressure variation must be approximately zero.

These conditions still allow the fundamental mode at approximately 41.7 Hz and the next mode at 125 Hz.

In fact, the midpoint is not a pressure node for either of those two modes, so the missing first-return pulse does not imply that a pressure microphone there must detect zero amplitude during resonance.

The exact constraint that changes the result is superposition: the microphone records the sum of pressure variations, not a separate count of how many reflected waves are present.

This reveals a useful distinction between two acoustic measurements.

A short pulse and its return time can tell us about distance and propagation speed. A sustained tone can reveal resonant frequencies, nodes, antinodes, and boundary conditions.

Both experiments depend on reflection, but they extract different information from the same travelling sound waves.