How to Find Beat Frequency When One Source Changes Pitch

Two sound sources are playing together. One produces a steady \(425\text{ Hz}\) tone. The other starts at \(415\text{ Hz}\), then gradually increases its pitch.

At first, you hear about ten loudness pulses each second. The pulses slow down, disappear, and then return.

Has the changing source slowed down?

No. Its frequency has increased the entire time. The pulses slow down because the two frequencies are getting closer together. Once the changing source passes \(425\text{ Hz}\), the difference starts increasing again, so the beats become faster.

The method is straightforward: subtract the two frequencies and take the absolute value.

If one source has a fixed frequency \(f_r\), and the other has a changing frequency \(f_s(t)\), the instantaneous beat frequency is

\[
\boxed{f_b(t)=|f_s(t)-f_r|}
\]

Here:

• \(f_b(t)\) is the beat frequency at time \(t\), in hertz.
• \(f_s(t)\) is the changing source’s frequency, in hertz.
• \(f_r\) is the fixed reference frequency, in hertz.

The important detail is that you must recalculate the difference as the source changes pitch. You cannot assume the beat frequency stays constant.

Let’s see why this works, using an adjustable air column as our changing sound source.

01The experiment: one fixed tone, one changing tone

Imagine an experiment with two sound sources.

The first is a stationary speaker producing a constant \(425\text{ Hz}\) tone.

The second is an adjustable pipe. It is open at one end and closed by a movable piston at the other. Blowing across its opening produces a resonating air column. Moving the piston changes the length of the air column and therefore changes its resonant frequency.

A microphone is positioned where sound from both sources overlaps. It records the combined pressure variations in the air.

Before calculating anything, make a prediction.

If the pipe is initially producing a frequency below \(425\text{ Hz}\), what happens to the beats when you shorten the pipe?

The pipe’s pitch increases. Initially, this brings its frequency closer to \(425\text{ Hz}\), so the beats become slower. When the pipe reaches \(425\text{ Hz}\), the beat frequency becomes zero. If you keep shortening the pipe, its frequency moves above \(425\text{ Hz}\), and the beats become faster again.

To understand why the pipe changes pitch, we need to look at what sound is doing inside it.

What actually oscillates in a sound wave?

Sound travelling through air is a longitudinal wave.

This means individual air particles oscillate back and forth parallel to the direction in which the sound travels.

They do not travel all the way from the speaker to the microphone.

Instead, each particle moves around its equilibrium position, transferring energy to neighbouring particles.

A compression is a region where air particles are closer together, producing higher pressure.

A rarefaction is a region where particles are further apart, producing lower pressure.

When sound from two sources reaches the same location, their pressure variations combine through superposition.

Sometimes the compressions reinforce each other. The resultant pressure variation becomes larger.

At other times, a compression from one wave overlaps with a rarefaction from the other. The waves partly cancel, reducing the resultant pressure variation.

This produces the changing loudness we recognise as beats.

Notice the distinction:

• Each source can maintain a constant amplitude.
• The frequency of one source can change.
• The resultant amplitude at the microphone rises and falls at a changing rate.

The air particles are still oscillating rapidly, at frequencies around hundreds of hertz. The beats describe the much slower change in the amplitude of the combined sound.

Why the pipe’s pitch changes: standing-wave boundary conditions

Inside the adjustable pipe, sound waves reflect from the ends of the air column.

The incident and reflected waves superpose to produce a standing wave.

A standing wave has positions where the oscillation is always zero, called nodes, and positions where the oscillation is greatest, called antinodes.

The positions of these nodes and antinodes depend on the boundaries of the system.

For our pipe, one end is closed by a rigid piston and the other is open.

At the closed end, air cannot move through the piston. This creates a displacement node.

At the open end, the air pressure remains approximately atmospheric. This creates a pressure node and a displacement antinode.

Remember that displacement and pressure have opposite boundary patterns:

For the fundamental frequency, the distance between the displacement node and antinode is one-quarter of a wavelength.

Therefore,

\[
L_{\text{eff}}=\frac{\lambda}{4}
\]

Rearranging,

\[
\lambda=4L_{\text{eff}}
\]

We also know the wave equation:

\[
v=f\lambda
\]

Substituting the wavelength gives

\[
\boxed{f_1=\frac{v}{4L_{\text{eff}}}}
\]

Here:

• \(f_1\) is the fundamental frequency, in hertz.
• \(v\) is the speed of sound in air, in metres per second.
• \(L_{\text{eff}}\) is the effective length of the air column, in metres.

This equation gives us the reason the pipe changes pitch.

A shorter air column produces a higher fundamental frequency. A longer air column produces a lower fundamental frequency.

For the calculations below, we will treat the given length as the effective length. In a real pipe, the displacement antinode sits slightly beyond the open end, so an end correction may be needed.

The standing-wave boundary conditions determine which wavelengths are allowed. Changing the pipe length changes those allowed wavelengths, which changes the resonant frequency.

02How to calculate beat frequency as the pitch changes

Now we can put the two ideas together.

Suppose the reference speaker remains at \(425\text{ Hz}\), while the adjustable pipe becomes progressively shorter.

Take the speed of sound as \(340\text{ m s}^{-1}\).

We can calculate the pipe’s fundamental frequency using

\[
f_s=\frac{340}{4L_{\text{eff}}}
\]

Then compare it with the reference frequency:

\[
f_b=|f_s-425|
\]

Illustrative data: shortening the air column

The following values are calculated using the ideal closed-open pipe model. They are illustrative data, not measurements from an actual experiment.

Look carefully at the results.

The pipe’s frequency increases continuously as its length decreases.

However, the beat frequency does not increase continuously.

It decreases from \(10.4\text{ Hz}\) to zero, then increases to \(4.3\text{ Hz}\).

This is the central idea when one source changes pitch:

The beat frequency depends on how far apart the two frequencies are, not on whether either frequency is increasing or decreasing.

A useful decision process

When you encounter one changing source in an HSC question, follow this sequence.

Step 1: Identify the fixed reference frequency.

Step 2: Determine the changing source’s frequency at the time you’re interested in. Use the supplied data, a graph, or an appropriate physical relationship.

Step 3: Calculate the absolute difference:

\[
f_b=|f_s-f_r|
\]

Step 4: If the question asks for the time between successive loudness maxima, calculate the beat period:

\[
T_b=\frac{1}{f_b}
\]

Step 5: Check whether the source is moving towards or away from the reference frequency.

If its frequency is moving towards the reference, the beats slow down. If it moves away from the reference, the beats speed up.

This last check is particularly useful when a question asks you to explain a changing pattern rather than simply calculate a number.

03When the changing frequency makes the calculation more complicated

The absolute-difference formula still works when one source changes pitch, but there is an important limitation.

If the source changes frequency slowly, you can calculate the instantaneous beat frequency using its current frequency.

If the frequency changes rapidly during the time between successive loudness maxima, however, there may not be one constant beat period for the whole observation.

In that situation, it helps to think about the relative phase of the two waves.

Phase describes where a wave is within its oscillation cycle.

When two waves have the same frequency, their phase difference remains constant.

When their frequencies differ, their relative phase changes. The larger the frequency difference, the faster that phase relationship changes.

Define the signed frequency difference as

\[
\Delta f(t)=f_s(t)-f_r
\]

The sign tells us which source has the higher frequency.

• If \(\Delta f\) is negative, the changing source has the lower frequency.
• If \(\Delta f\) is positive, the changing source has the higher frequency.
• If \(\Delta f=0\), the frequencies are equal.

The beat frequency is the magnitude of this difference:

\[
f_b(t)=|\Delta f(t)|
\]

There is a useful extension when the frequency changes continuously.

The signed area under a graph of \(\Delta f\) against time gives the change in relative phase, measured in cycles.

This is because frequency measures cycles per second. Multiplying by a time interval gives a number of cycles.

For a changing frequency, you can therefore use the area of rectangles or triangles under a frequency-difference graph to track how far the waves move in and out of phase.

We’ll use this idea in the final question.

A common misconception: zero beats means silence

Suppose two sources produce exactly the same frequency.

Their relative phase no longer changes, so there are no beats.

But does that mean the sound disappears?

Not necessarily.

If the waves are in phase, their amplitudes reinforce each other continuously. The resultant sound has a constant, large amplitude.

If they are exactly out of phase and have equal amplitudes, they cancel at the observation point.

If their amplitudes are unequal, there is a constant resultant amplitude rather than complete cancellation.

In all three cases, the beat frequency is zero.

Zero beat frequency means there is no changing amplitude pattern. It does not automatically mean zero sound.

This distinction becomes particularly important when one source passes through the reference frequency. The phase relationship at the crossing depends on what happened before it.

Pitch changes, volume changes, and Doppler shifts

These three situations can look similar in a question, but they describe different physics.

For a stationary adjustable pipe, the emitted frequency changes because its resonant air-column length changes.

For a moving source, the emitted frequency might stay constant while the observer receives a different frequency.

In that second situation, calculate the Doppler-shifted frequency first, then compare it with the reference frequency.

If you need to revise that distinction, see The Doppler Effect: Frequency Changes in HSC Physics.

For a fuller explanation of why two nearby frequencies produce beats in the first place, see Beat Frequency in HSC Physics: How Beats Work.

04Worked examples

05Questions and solutions

Question 1

Two stationary speakers produce sound simultaneously.

Speaker A produces a constant frequency of \(512\text{ Hz}\). Speaker B currently produces \(507\text{ Hz}\).

Assuming Speaker B’s frequency changes slowly enough to remain approximately constant during each beat, calculate:

a) The instantaneous beat frequency.

b) The time between successive loudness maxima.

Solution 1

The beat frequency is \(5.0\text{ Hz}\), and successive loudness maxima are separated by \(0.20\text{ s}\).

Part a: Beat frequency

Use

\[
f_b=|f_1-f_2|
\]

Substituting,

\[
\begin{aligned}
f_b&=|512-507|\\
&=5.0\text{ Hz}
\end{aligned}
\]

Part b: Beat period

The beat period is the reciprocal of the beat frequency:

\[
T_b=\frac{1}{f_b}
\]

Therefore,

\[
T_b=\frac{1}{5.0}
=0.20\text{ s}
\]

The combined sound reaches a loudness maximum every \(0.20\text{ s}\).

Remember that this is the period of the beat pattern, not the period of either individual sound wave.

Question 2

A closed-open pipe has a movable piston and is blown across its open end.

Its effective air-column length is initially \(0.205\text{ m}\). A stationary reference speaker produces a frequency of \(425\text{ Hz}\).

Take the speed of sound as \(340\text{ m s}^{-1}\).

a) Explain why the closed end is a displacement node and the open end is a displacement antinode.

b) Calculate the fundamental frequency of the pipe.

c) Calculate the initial beat frequency.

d) The piston is moved until the effective length becomes \(0.200\text{ m}\). Calculate the new beat frequency and explain what happens to the loudness pattern.

Assume the fundamental mode and neglect end correction.

Solution 2

The initial pipe frequency is approximately \(414.6\text{ Hz}\), giving an initial beat frequency of \(10.4\text{ Hz}\). After shortening the pipe to \(0.200\text{ m}\), the beat frequency becomes zero.

Part a: Boundary conditions

At the closed end, the rigid piston prevents air particles from moving through the boundary. Therefore, the air-displacement amplitude is zero at that point, creating a displacement node.

At the open end, the air pressure can equalise with atmospheric pressure. This creates a pressure node and a displacement antinode.

These boundary conditions determine the wavelengths that can form standing waves inside the pipe.

Part b: Fundamental frequency

For a closed-open pipe,

\[
f_1=\frac{v}{4L_{\text{eff}}}
\]

Substituting,

\[
\begin{aligned}
f_1&=\frac{340}{4(0.205)}\\
&=\frac{340}{0.820}\\
&=414.63\text{ Hz}
\end{aligned}
\]

Therefore,

\[
f_1\approx414.6\text{ Hz}
\]

Part c: Initial beat frequency

\[
f_b=|f_1-f_r|
\]

\[
\begin{aligned}
f_b&=|414.63-425|\\
&=10.37\text{ Hz}
\end{aligned}
\]

The initial beat frequency is approximately \(10.4\text{ Hz}\).

Part d: New beat frequency

The new effective length is \(0.200\text{ m}\).

\[
\begin{aligned}
f_2&=\frac{340}{4(0.200)}\\
&=\frac{340}{0.800}\\
&=425\text{ Hz}
\end{aligned}
\]

The new beat frequency is

\[
\begin{aligned}
f_b&=|425-425|\\
&=0\text{ Hz}
\end{aligned}
\]

The two frequencies are now equal.

As the pipe is shortened from its initial length, its frequency increases towards the reference frequency. The beats therefore become progressively slower until the two frequencies match.

At that point, the relative phase stops changing, so the loudness no longer oscillates periodically.

The sound does not necessarily disappear. Its resultant amplitude remains constant if the frequencies remain equal.

Question 3

A fixed speaker produces a frequency of \(500\text{ Hz}\).

A second electronic oscillator starts at \(492\text{ Hz}\) and increases its frequency uniformly to \(508\text{ Hz}\) over \(4.0\text{ s}\).

a) Calculate the oscillator’s frequency after \(1.0\text{ s}\).

b) Calculate the instantaneous beat frequency at this time.

c) Repeat parts a) and b) after \(3.0\text{ s}\).

d) Determine when the beat frequency becomes zero.

e) Describe how the beat frequency changes throughout the four-second interval.

Solution 3

The instantaneous beat frequency is \(4.0\text{ Hz}\) at both \(1.0\text{ s}\) and \(3.0\text{ s}\). The beats disappear at \(2.0\text{ s}\).

Part a: Frequency after one second

The frequency increases uniformly.

Its rate of increase is

\[
\frac{508-492}{4.0}
=4.0\text{ Hz s}^{-1}
\]

After \(1.0\text{ s}\), the frequency has increased by

\[
\Delta f=4.0(1.0)=4.0\text{ Hz}
\]

Therefore,

\[
f_s=492+4=496\text{ Hz}
\]

Part b: Beat frequency after one second

\[
\begin{aligned}
f_b&=|f_s-f_r|\\
&=|496-500|\\
&=4.0\text{ Hz}
\end{aligned}
\]

Part c: Frequency and beat frequency after three seconds

After \(3.0\text{ s}\), the oscillator’s frequency has increased by

\[
\Delta f=4.0(3.0)=12\text{ Hz}
\]

Therefore,

\[
f_s=492+12=504\text{ Hz}
\]

The beat frequency is

\[
\begin{aligned}
f_b&=|504-500|\\
&=4.0\text{ Hz}
\end{aligned}
\]

The beat frequency is again \(4.0\text{ Hz}\).

Part d: When do the beats disappear?

The beats disappear when the oscillator reaches \(500\text{ Hz}\).

It must increase by

\[
500-492=8\text{ Hz}
\]

Since it increases at \(4.0\text{ Hz s}^{-1}\),

\[
t=\frac{8}{4.0}=2.0\text{ s}
\]

Therefore, the beat frequency reaches zero after \(2.0\text{ s}\).

Part e: Overall behaviour

At the beginning, the frequency difference is

\[
|492-500|=8\text{ Hz}
\]

The difference decreases as the oscillator approaches \(500\text{ Hz}\).

At \(2.0\text{ s}\), the frequencies match and the beat frequency becomes zero.

After this point, the oscillator continues increasing in frequency. It moves further above \(500\text{ Hz}\), so the beat frequency increases again.

The interesting feature is that the beat frequency has the same value at \(1.0\text{ s}\) and \(3.0\text{ s}\), but it is moving in opposite directions at those times.

The first measurement is on the way towards the reference frequency. The second is on the way away from it.

Question 4

A stationary speaker produces a reference frequency of \(700\text{ Hz}\).

A second source emits a frequency of \(680\text{ Hz}\) while moving directly towards a stationary observer at \(10\text{ m s}^{-1}\).

The speed of sound in air is \(340\text{ m s}^{-1}\).

Calculate:

a) The frequency of the moving source as received by the observer.

b) The beat frequency between the moving source and the stationary speaker.

c) Explain why calculating the beat frequency using the emitted frequency of \(680\text{ Hz}\) gives an incorrect result.

Assume the reference speaker and observer are stationary.

Solution 4

The moving source is received at approximately \(700.6\text{ Hz}\), giving a beat frequency of approximately \(0.61\text{ Hz}\).

Part a: Observed frequency

Because the source moves towards a stationary observer, we use the Doppler equation:

\[
f_{\text{obs}}=f_s\frac{v}{v-v_s}
\]

where:

• \(f_{\text{obs}}\) is the observed frequency.
• \(f_s\) is the emitted frequency.
• \(v\) is the speed of sound.
• \(v_s\) is the source speed towards the observer.

Substituting,

\[
\begin{aligned}
f_{\text{obs}}
&=680\left(\frac{340}{340-10}\right)\\
&=680\left(\frac{340}{330}\right)\\
&=700.61\text{ Hz}
\end{aligned}
\]

Therefore,

\[
f_{\text{obs}}\approx700.6\text{ Hz}
\]

Part b: Beat frequency

The stationary reference speaker is received at its original frequency of \(700\text{ Hz}\).

We therefore compare the two frequencies at the observer:

\[
\begin{aligned}
f_b&=|f_{\text{obs}}-f_r|\\
&=|700.61-700|\\
&=0.61\text{ Hz}
\end{aligned}
\]

The beat frequency is approximately \(0.61\text{ Hz}\).

Part c: Why the emitted frequency is not enough

A tempting calculation is

\[
|700-680|=20\text{ Hz}
\]

However, this ignores the Doppler effect.

The moving source emits \(680\) cycles per second, but its movement compresses the wavefronts in front of it. The stationary observer therefore encounters those wavefronts more frequently.

The observer receives approximately \(700.6\) cycles per second, not \(680\).

Beat frequency depends on the frequencies received at the same observation point, so the Doppler-shifted frequency must be used.

This is a useful example of two wave concepts working together: Doppler shift determines the received frequency, while interference determines the beat frequency.

Question 5

Two sound sources produce waves of equal amplitude at a stationary microphone.

Source A maintains a constant frequency of \(500\text{ Hz}\).

Source B starts at \(496\text{ Hz}\) and increases its frequency uniformly to \(504\text{ Hz}\) over \(4.0\text{ s}\).

At \(t=0\), the two waves are exactly in phase at the microphone.

Assume:

• Both sources and the microphone remain stationary.
• The amplitudes received from both sources remain equal.
• The frequency of Source B increases linearly.
• The phase of each wave changes continuously, with no sudden phase shift.

Answer the following:

a) At what time does the instantaneous beat frequency become zero?

b) Calculate the change in relative phase, measured in cycles, between \(t=0\) and the instant found in part a).

c) At that instant, are the waves in phase or out of phase?

d) Is the resultant sound at a maximum or minimum amplitude at that instant?

e) How many complete cycles of relative-phase change occur over the full four seconds, counting changes in either direction?

Solution 5

The beat frequency becomes zero at \(2.0\text{ s}\). At that instant, the waves are back in phase, so their resultant amplitude is at a maximum. Over the full four seconds, they undergo eight complete cycles of relative-phase change.

This question requires us to distinguish two things:

1. How quickly the relative phase is changing.
2. What the relative phase actually is at a particular instant.

The first determines the beat frequency. The second determines whether the waves reinforce or cancel.

Part a: Time when the beat frequency becomes zero

Source B increases from \(496\text{ Hz}\) to \(504\text{ Hz}\).

Its rate of increase is

\[
\frac{504-496}{4.0}
=2.0\text{ Hz s}^{-1}
\]

The frequency difference is initially \(4\text{ Hz}\), with Source B below Source A.

It must increase by \(4\text{ Hz}\) to match Source A.

Therefore,

\[
t=\frac{4}{2.0}
=2.0\text{ s}
\]

The instantaneous beat frequency becomes zero at \(2.0\text{ s}\).

Part b: Change in relative phase

This is where simply substituting the instantaneous frequencies is not enough.

We need to determine how much relative phase has accumulated between \(t=0\) and \(t=2.0\text{ s}\).

The signed frequency difference is initially

\[
\Delta f=-4\text{ Hz}
\]

It increases uniformly to

\[
\Delta f=0\text{ Hz}
\]

over two seconds.

The change in relative phase, measured in cycles, is the signed area under the frequency-difference-versus-time graph.

This region is a triangle.

Its area is

\[
\begin{aligned}
N&=-\frac{1}{2}\times
\text{base}\times\text{height}\\
&=-\frac{1}{2}\times2.0\times4\\
&=-4\text{ cycles}
\end{aligned}
\]

The negative sign means Source B has fallen behind Source A by four complete cycles.

Part c: Are the waves in phase or out of phase?

They are back in phase.

At the beginning, the waves are exactly in phase.

Over the next two seconds, their relative phase changes by negative four complete cycles.

A complete cycle corresponds to a phase change of \(360^\circ\), or \(2\pi\) radians.

Therefore, the total phase change is

\[
-4\times360^\circ=-1440^\circ
\]

This is four complete rotations of relative phase.

After four complete cycles, the waves return to their original phase relationship.

So, at \(t=2.0\text{ s}\), they are in phase again.

Part d: Resultant amplitude

The resultant amplitude is at a maximum.

Both waves have equal amplitudes and are in phase, so they reinforce each other through constructive interference.

If each individual wave has amplitude \(A\), the resultant amplitude is

\[
A_{\text{resultant}}=2A
\]

This exposes a particularly tempting misconception.

A student might argue that because the beat frequency is zero, the sound must be at its quietest point.

But zero beat frequency only tells us that the relative phase is momentarily no longer changing.

It does not tell us what that phase actually is.

In this question, the waves happen to be in phase, so the resultant amplitude is at its maximum.

Part e: Total relative-phase cycles

We must now consider the second half of the frequency sweep.

Between \(t=2.0\text{ s}\) and \(t=4.0\text{ s}\), Source B continues increasing from \(500\text{ Hz}\) to \(504\text{ Hz}\).

The signed frequency difference increases from zero to \(+4\text{ Hz}\).

The relative phase therefore changes by another four cycles, but this time in the positive direction.

The total phase movement is:

• First two seconds: four cycles in the negative direction.
• Last two seconds: four cycles in the positive direction.

Therefore, the total number of complete relative-phase cycles, counting both directions, is

\[
N_{\text{total}}=4+4
\]

\[
N_{\text{total}}=8\text{ cycles}
\]

There is one final detail worth noticing.

The signed phase change over the entire four seconds is actually zero, because the negative and positive changes cancel.

However, the waves have still moved through eight complete cycles of relative phase.

This distinction between signed phase change and total phase movement is useful whenever a frequency difference changes direction.

06Where to go next

When one sound source changes pitch, the key is to keep track of the frequency difference rather than treating the beat frequency as a fixed value.

For a slowly changing source, use

\[
\boxed{f_b(t)=|f_s(t)-f_r|}
\]

If the source passes through the reference frequency, expect the beat rate to decrease to zero and then increase again.

The next useful connection is to standing-wave resonance and the Doppler effect. Standing-wave boundary conditions explain how a source such as an air column produces a particular pitch. The Doppler effect explains how the pitch received by an observer can differ from the frequency actually emitted.

Together, these ideas let you distinguish a source that changes its own frequency from one that simply moves relative to the listener.