Case 2: observer at rest, source moving
Now consider the case where the observer is at rest and the source moves. This situation is fundamentally different from Case 1. The motion of the source changes when and where consecutive maxima are emitted. We will calculate the time interval between the arrivals of two consecutive compression maxima at the observer.
Derivation: source approaching the observer
Let the source emit maxima at frequency \(f\), with a period \(T = 1/f\) between consecutive emissions. Suppose that at time \(t = 0\) the source is at a distance \(x\) from the observer at rest, and emits the first maximum. Fig 4.1 shows this for a source at rest, as a baseline.
The first maximum travels at speed \(c\), so it reaches the observer at the time
\[ t_1 = \dfrac{x}{c} \]
At time \(t = T\), the source emits the second maximum. But during this interval \(T\), the source has moved a distance \(v_S T\) closer to the observer. So the second maximum is emitted from a distance \(x - v_S T\), as Fig 4.2 shows.
The second maximum travels this reduced distance at speed \(c\). It was emitted at time \(T\), so it reaches the observer at the time
\[ t_2 = T + \dfrac{x - v_S T}{c} \]
The observer perceives a new period \(T'\), the difference of the two arrival times. The derivation below finds it.
\[ T' = t_2 - t_1 = \left(T + \dfrac{x - v_S T}{c}\right) - \dfrac{x}{c} \]
Why. the observed period is the time between the two arrivals.
\[ T' = T + \dfrac{x}{c} - \dfrac{v_S T}{c} - \dfrac{x}{c} = T - \dfrac{v_S T}{c} \]
Why. open the bracket; the two terms \(x/c\) cancel, so the starting distance drops out.
\[ T' = T\left(1 - \dfrac{v_S}{c}\right) = T\left(\dfrac{c - v_S}{c}\right) \]
Why. take \(T\) out as a common factor.
\[ f' = \dfrac{1}{T'} = \dfrac{c}{(c - v_S)T} \]
Why. the apparent frequency is the reciprocal of the observed period.
\[ f' = \dfrac{c}{c - v_S} \cdot f \]
Read it. \(1/T = f\).
When the source moves towards an observer at rest, the frequency increases. The distance \(x\) cancelled in step 2. So the result does not depend on how far away the source is, only on its speed.
A source of \(450\,\text{Hz}\) moves straight towards an observer at rest at \(34\,\text{m/s}\). Take \(c = 340\,\text{m/s}\). What frequency does the observer hear, in Hz?
Source moving away
If the source moves away from the observer at speed \(v_S\), the second maximum is emitted from the distance \(x + v_S T\). Choose the second chip of Fig 4.2 to see it. Following the same derivation,
\[ t_2 = T + \dfrac{x + v_S T}{c} \]
\[ T' = t_2 - t_1 = T + \dfrac{v_S T}{c} = T\left(\dfrac{c + v_S}{c}\right) \]
\[ f' = f\left(\dfrac{c}{c + v_S}\right) \]
When the source moves away from an observer at rest, the frequency decreases.
An auto-rickshaw sounds a \(480\,\text{Hz}\) horn while it moves at \(20\,\text{m/s}\) along a straight road. A shopkeeper stands at the roadside ahead. Take \(c = 340\,\text{m/s}\). What frequency does the shopkeeper hear (a) as the auto approaches, and (b) after it has passed?
(a) Approaching: \(f' = f\,c/(c - v_S) = 480 \times 340/320 = 510\,\text{Hz}\).
(b) Receding: \(f' = f\,c/(c + v_S) = 480 \times 340/360 \approx 453.3\,\text{Hz}\).
The rise, \(30\,\text{Hz}\), is larger than the fall, \(26.7\,\text{Hz}\). The factors \(c/(c - v_S)\) and \(c/(c + v_S)\) are not symmetric.
A \(1000\,\text{Hz}\) source and an observer approach each other at \(20\,\text{m/s}\) in two separate experiments. In one, only the source moves. In the other, only the observer moves. Take \(c = 340\,\text{m/s}\). Which statement is correct?
The general Doppler formula
Combining the two cases gives a single equation. It covers the motion of both the source and the observer along the line connecting them:
\[ f' = f\left(\dfrac{c \pm v_O}{c \mp v_S}\right) \]
Here \(f'\) is the apparent (observed) frequency and \(f\) the actual frequency of the source. \(c\) is the speed of sound in the medium. \(v_O\) is the speed of the observer and \(v_S\) the speed of the source.
The sign convention
The signs in the formula depend on the directions of motion. A simple rule is easy to remember. Motion that brings the source and the observer closer increases the frequency. Motion that separates them decreases the frequency.
Numerator (observer, \(v_O\)). If the observer moves towards the source, use \(+\): \((c + v_O)\). If the observer moves away, use \(-\): \((c - v_O)\).
Denominator (source, \(v_S\)). If the source moves towards the observer, use \(-\): \((c - v_S)\). If the source moves away, use \(+\): \((c + v_S)\).
One point is important. All velocities must be measured with respect to the medium, the air, not relative to each other. The speed of sound \(c\) is its speed in the air, so \(v_O\) and \(v_S\) must be speeds in the air too.
Car A sounds a \(480\,\text{Hz}\) horn and drives at \(20\,\text{m/s}\). Car B drives at \(10\,\text{m/s}\) straight towards car A on the same road. Take \(c = 340\,\text{m/s}\) in still air. What frequency does the driver of B hear?
Both motions bring the cars closer, so the observer term takes \(+\) and the source term takes \(-\).
\(f' = f(c + v_O)/(c - v_S) = 480 \times (340 + 10)/(340 - 20) = 480 \times 350/320 = 525\,\text{Hz}\).
A source moves away from an observer, and the observer moves towards the source. Which form of the formula applies?
Motion at an angle
If the motion is at an angle, use only the component of the velocity along the line connecting the source and the observer. The perpendicular component neither brings the two closer nor takes them apart. So it does not change the rate at which maxima arrive.
In the figure below, a police car with an \(800\,\text{Hz}\) siren drives at \(v = 20\,\text{m/s}\) under a footbridge. A listener O stands on the bridge, \(20\,\text{m}\) above the siren. With the car at a distance \(x\) along the road, the line to O makes an angle with the road. The component of \(v\) along that line is \(v_{\parallel} = v\,|x|/\sqrt{x^2 + 20^2}\).
The pitch does not jump. It slides down smoothly as the car passes, because the component along the line shrinks to zero and then grows the other way. Far from the bridge, the line is nearly along the road, and the whole \(20\,\text{m/s}\) counts. Then the frequency approaches \(800 \times 340/320 = 850\,\text{Hz}\) before and \(800 \times 340/360 \approx 755.6\,\text{Hz}\) after.
In the figure above, the car is exactly under the listener. What frequency does the listener hear at that instant?
A source and an observer are on a straight road in still air. Which of these raise the frequency that the observer hears above the source frequency? Select all that apply.
In 1845 the Dutch scientist Christophorus Buys Ballot tested Doppler's idea with sound. He put horn players on an open carriage of a train near Utrecht, and asked them to hold one steady note. Musicians with a good ear stood beside the track and named the note they heard. As the train approached, they heard it higher, and as it moved away, lower. The size of the change agreed with Doppler's formula.
- Putting the source's speed in the numerator.The observer's speed goes in the numerator and the source's speed in the denominator. They come from different derivations.
- Using the relative speed of the source and the observer in one formula.Measure each speed relative to the medium, the air, and put each in its own place.
- Using the full speed of a vehicle that moves at an angle to the line of sight.Resolve the velocity. Only the component along the line joining source and observer changes the frequency.