Doppler Effect IB DP Physics Study Notes

Doppler effect IB DP Physics Study Notes

Doppler effect IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand

the nature of the Doppler effect for sound waves and electromagnetic waves
the representation of the Doppler effect in terms of wavefront diagrams when either the source or the observer is moving
the relative change in frequency or wavelength observed for a light wave due to the Doppler effect where the speed of light is much larger than the relative speed between the source and the observer
as given by $\frac{\Delta f}{f} = \frac{\Delta \lambda}{\lambda} \approx \frac{v}{c}$
that shifts in spectral lines provide information about the motion of bodies like stars and galaxies in
space
The observed frequency for sound waves and mechanical waves due to the Doppler effect is given by
Moving Source:
$f’ = f \frac{v}{v \pm u_s}$
where $u_s$ is the velocity of the source.
Moving Observer:
$f’ = f \frac{v \pm u_o}{v}$
where $u_o$ is the velocity of the observer.

Standard level and higher level: 2 hours
Additional higher level: 2 hours

IB DP Physics 2025 -Study Notes -All Topics

The Doppler effect – moving source

The Doppler effect explains frequency change caused by moving sources or moving observers.
For example, if a sound source of frequency f approaches you at a speed uS, its wavefronts will bunch togetherand you will hear a frequency f ’ which is higher than f. ( f ’ > f ).
On the other hand, if a sound source recedes from you at a speed uS, its wavefronts will stretch out and you will hear a frequency f ’ which is lower than f. ( f ’ < f ).
Consider the ice-cream truck that parks (u_S = 0) in the neighborhood playing that interminable music.
Since the ice cream truck is not yet moving, the wave fronts are spherically symmetric.

The Doppler effect – moving source

Whether Dobson is in front of the truck, or behind it, he hears the same frequency.

FYI

Even though the sound waves are drawn as transverse, they are actually longitudinal.

∙Suppose the ice-cream truck is now moving, but ringing the bell at the same rate as before.
∙Note how the wave fronts bunch up in the front, and separate in the back.

∙The reason this happens is that the truck moves forward a little bit during each successive spherical wave emission.

FYI

∙Don’t forget, the actual speed of the wavefronts through the stationary medium of the air is the speed of sound v (bunched or otherwise).

Dobson will hear a different frequency now, depending on his position relative to the truck.
If Dobson is in front of the moving truck, f ’ > f.
If he is behind, he will hear f ’ < f.

Now we want to look at TWO successive pulses emitted by a source that is MOVING at speed uS:
λ is the “real” wavelength of the sound.
Since the speed of sound in air is v, λ = vT where T is the period of the sound source. Note that Pulse 1 has traveled λ = vT in the time t = T.
λ’ is the wavelength detected by the observer.
The distance d between the emission of the two pulses is simply the velocity of the source u_S times the time between emissions T. Thus d = u_ST.
Since λ = d + λ’ then λ’ = λ – d = vT – u_ST. So
λ’ = (v – u_S)T
From $v = \lambda’ f’$ and $\lambda’ = (v – u_s) T$, we see that:
\[f’ = \frac{v}{\lambda’} = \frac{v}{(v – u_s)T}\]
\[f’ = \left(\frac{1}{T}\right) \left(\frac{v}{v – u_s}\right) \quad \text{[Since $f = \frac{1}{T}$]}\]
\[f’ = f \left(\frac{v}{v – u_s}\right)\]
FYI: If the source is receding, replace $u_s$ with $-u_s$.’

The Doppler effect – moving observer

Suppose the source is stationary.
If the observer is not moving, he will hear the true frequency of the source.
The diagram shows which wavefronts pass the observer in the time interval $\Delta t$ (red):
Because the wave speed is $v$, the length of the region in question is $v \Delta t$.
If we divide the length $v \Delta t$ by the wavelength $\lambda$, we get the number of cycles (or wavefronts). Thus:

\[\#\text{cycles detected} = \frac{v \Delta t}{\lambda}.\]

From the definition of $f’$, we have:

\[f’ = \frac{\#\text{cycles detected}}{\Delta t} = \frac{\left( \frac{v \Delta t}{\lambda} \right)}{\Delta t} = \frac{v}{\lambda}.\]

But $\frac{v}{\lambda} = f$, so that $f’ = f$. (EXPECTED)

The diagram shows which wavefronts pass the moving observer in the time interval Δt.
The red area $v\Delta t$ is still due to the wave speed itself.
The blue area $u_0\Delta t$ is due to the observer’s speed $u_0$.
Now for the observer

#cycles detected = $\frac{(v\Delta t + u_0\Delta t)}{\lambda}$.
$f’ = \frac{\#cycles detected}{\Delta t}$
= $\frac{(v\Delta t + u_0\Delta t)}{\lambda\Delta t}$
= $\frac{v}{\lambda} + \frac{u_0}{\lambda}$

From $f’ = \frac{v}{\lambda} + \frac{u_o}{\lambda}$ and the relation $v = f\lambda$ we have

$f’ = \frac{v}{\lambda} + \frac{u_o}{\lambda}$

$f’ = f + \left(\frac{u_o}{\lambda}\right)\left(\frac{v}{f}\right)$

$f’ = \frac{fv}{v} + \frac{fu_o}{v}$

$f’ = f \left(\frac{v+u_o}{v}\right)$.

If the observer is moving AWAY from the source, substitute $-u_o$ for $u_o$.

FYI: Always pick the sign that makes f ’ do what is expected.

The Doppler effect – light

As light is a wave, it can also undergo these Doppler shifts. The derivation is complicated but we can calculate the change in frequency using the formula:

$\Delta f = \frac{v}{c}f$

where:

$\Delta f$ – change in frequency

$v$ – relative speed between source and observer

$c$ – speed of light

$f$ – original frequency at the source

More completely the relationships are

Situations where the Doppler effect can be utilized

The radar gun used by police. It uses the form $v = \frac{cΔf}{f}$ to find the velocity of the car. It actually measures the difference in frequency between the emitted radar beam, and the reflected-and-returned on

A Doppler ultrasound test uses reflected sound waves to see how blood flows through a blood vessel. It helps doctors evaluate blood flow through major arteries and veins, such as those of the arms, legs, and neck. It can show blocked or reduced blood flow through narrowing in the major arteries of the neck that could cause a stroke. It also can reveal blood clots in leg veins (deep vein thrombosis, or DVT) that could break loose and block blood flow to the lungs (pulmonary embolism).

A Doppler weather radar is a specialized radar that makes use of the Doppler effect to produce velocity data about objects at a distance. It does this by beaming a microwave signal towards a desired target and listening for its reflection, then analyzing how the frequency of the returned signal has been altered by the storm’s motion. This variation gives accurate measurements of the radial component of a storm’s velocity.

Question

Source S produces sound waves of speed $v$ and frequency $f$. S moves with constant velocity $\frac{v}{5}$ away from a stationary observer.

What is the frequency measured by the observer?

A. $\frac{4}{5} f$

B. $\frac{5}{6} f$

C. $\frac{6}{5} f$

D. $\frac{5}{4} f$

▶️Answer/Explanation

Ans:B

The frequency observed by a stationary observer when a source is in motion can be calculated using the Doppler effect formula for sound:

\[f’ = \frac{f \cdot (v + v_o)}{v + v_s},\]

$f$ is the frequency emitted by the source.
$v$ is the speed of sound.
$v_o$ is the velocity of the observer, which is stationary (0).
$v_s$ is the velocity of the source, which is moving away at $\frac{v}{5}$.

Now, substitute these values into the formula:

\[f’ = \frac{f \cdot (v + 0)}{v + \frac{v}{5}}.\]

Simplify the equation:

\[f’ = \frac{f \cdot v}{\frac{6}{5}v} = \frac{5}{6}f.\]).

Question

A radar detector is used to measure the speed of a car. The car is moving with a speed v towards the detector.

The detector emits microwaves of frequency f and speed c. Which of the following is the change in frequency of the microwaves measured at the detector after reflection by the car?

▶️Answer/Explanation

Ans C

Doppler Effect IB DP Physics Study Notes