A point A oscillates according to a certain harmonic law in the reference frame K′ which in its turn performs harmonic oscillations relative to the reference frame K. Both oscillations occur along the same direction. When the K′ frame oscillates at the frequency 20 or 24 Hz, the beat frequency of the point A in the K frame turns out to be equal to ν. At what frequency of oscillation of the frame K′ will the beat frequency of the point A become equal to 2ν?
For the given numbers the frame K′ must oscillate at 18 Hz or at 26 Hz. With the sliders as they are now, the beat frequency is ν = Hz, the frequency of A in K′ is ν0 = Hz, and a beat of ν needs ν̄ = Hz or Hz.
Problem statement
Problem 4.13. A point A oscillates according to a certain harmonic law in the reference frame K′ which in its turn performs harmonic oscillations relative to the reference frame K. Both oscillations occur along the same direction. When the K′ frame oscillates at the frequency 20 or 24 Hz, the beat frequency of the point A in the K frame turns out to be equal to ν. At what frequency of oscillation of the frame K′ will the beat frequency of the point A become equal to 2ν?
Interpretation. The point A oscillates along a line with frequency ν0 as seen in K′. The frame K′ itself oscillates along the same line, with frequency ν̄, as seen in K. In K the position of A is the sum of these two motions, so it has two frequencies and shows beats. The statement gives two values of ν̄, 20 Hz and 24 Hz, with the same beat frequency ν. The question asks for ν̄ when the beat is 2ν.
We assume the amplitude of A in K′ and the amplitude of K′ in K are 3 cm each. The problem does not give them. They change only the size of the waves and the depth of the beats, never a frequency. They are sliders. The frequencies 20 Hz and 24 Hz and the factor 2 are the ones in the problem, and they are sliders too.
Symbols: ν0 is the frequency of A in K′, ν̄ the frequency of K′ in K, ν the beat frequency in the first two cases, m the multiple of the beat asked for (2 in the problem).
The big idea in plain words
The point A has two motions at once: its own oscillation in K′, and the shaking of the whole frame K′. Seen from K, the two add. Two oscillations of different frequencies add to give beats, and the beat frequency is the difference of the two frequencies. So changing the frequency of the frame K′ changes the beat. Two frame frequencies, one above and one below ν0, give the same beat.
The solution in each frame
Frame K′ (moving with the oscillating frame). A oscillates at ν0 alone, so there are no beats here. Frame K (the fixed frame). Position of A in K = position of K′ in K + position of A in K′. This is the sum of two oscillations, at ν̄ and ν0, so the beats appear here. The page shows four cases of K in four rows: the two given frequencies and the two answers.
The solution step by step
Step 1, the beat in K. The position of A in K is x = a cos(2πν0t) + b cos(2πν̄t), where a and b are the two amplitudes. Two oscillations of the same direction and different frequencies give beats at the frequency |ν̄ − ν0|.
Step 2, use the given data. The beat is ν for ν̄ = Hz and also for ν̄ = Hz. So |ν̄ − ν0| = ν for both. The two values of ν̄ are different, so one must be above ν0 and the other below: ν1 = ν0 − ν and ν2 = ν0 + ν (or the other way round).
Step 3, solve for ν0 and ν. Add the two: ν0 = (ν1 + ν2)/2 = Hz. Subtract: ν = |ν2 − ν1|/2 = Hz. For 20 Hz and 24 Hz these are 22 Hz and 2 Hz.
Step 4, the new beat. A beat of mν = Hz needs |ν̄ − ν0| = mν, so ν̄ = ν0 ± mν. With m = 2: ν̄ = 22 − 4 = 18 Hz or 22 + 4 = 26 Hz. For the numbers on screen the answers are Hz and Hz.
Step 5, the cases. A frequency must be above zero. If ν0 − mν is zero or below, only the upper answer is possible. If the two given frequencies are equal, ν = 0 and there is no beat, so the question has no answer.
A separate calculation builds the waveform in K for each of the four cases over 10 s, scans its spectrum within 0.5 Hz of the two expected lines, and measures the gap between the two peaks it finds. It compares each gap with ν or mν. The largest relative difference is .
See it and hear it
The picture shows the frame K′ as a green rectangle that oscillates along the fixed axis of K. Inside it, A oscillates about the middle of the frame. A’s position in K is the frame’s position plus A’s position in the frame. The waves of the frame and of A are drawn on one axis, with their sum below. Where the two waves are in step the sum is large (the green bands marked adds). Where they are opposite the sum is small (the red bands marked cancels). Tick the sound box and press a button to hear the same pair of tones. Both tones are shifted up by 400 Hz so that they are audible, and the difference between them, which is the beat, is the real one.
Worked calculation in one table
The frequency of A in K′ is one fixed number, ν0, that the problem does not give. The frame K′ is tried at 20 Hz and then at 24 Hz. These are two separate trials, not two oscillations at the same time. In K the point is the sum of A’s motion and the frame’s motion, so the beat is |ν̄ − ν0|, where ν̄ is the frequency of K′.
| Case | Frequency of A in K′ | Frequency of K′ in K | Beat seen in K |
|---|---|---|---|
| Trial 1 | ν0 | 20 Hz | |20 − ν0| = ν |
| Trial 2 | ν0 | 24 Hz | |24 − ν0| = ν |
| Same beat, so | ν0 = (20 + 24)/2 = 22 Hz, and ν = |20 − 22| = 2 Hz | ||
| Lower answer | 22 Hz | 18 Hz | |18 − 22| = 4 Hz = 2ν |
| Upper answer | 22 Hz | 26 Hz | |26 − 22| = 4 Hz = 2ν |
Two trials are enough: one trial gives one equation with two unknowns (ν0 and ν), and the second trial makes it solvable. Any extra trial only repeats the rule |ν̄ − 22|.
Five methods
1. Force method
Not suited: no forces or masses are given.
2. Energy method
Not suited, for the same reason.
3. Graph method
The first graph shows the beat |ν̄ − ν0| against ν̄, a V with its tip at ν0. A horizontal line at height ν cuts the V twice, at 20 and 24 Hz, so the tip is half way between them at 22 Hz. A line at 2ν cuts it at 18 and 26 Hz.
4. Calculus method
Not needed. The only algebra is the two equations in step 2.
5. Check method
Units: all quantities are frequencies in Hz. Check the answer: for ν̄ = 18 and 26, the beats are |18 − 22| = 4 Hz and |26 − 22| = 4 Hz, which equal 2ν.
Fastest: the V graph. Most reliable: the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Frame K′ frequency in trial 1 and trial 2 | Yes | The frequency of A in K′ (ν0, the middle of the two), the beat, both answers, the traces and every graph |
| Multiple m | Yes | The two answers, the lower two traces and the graph of answers against m |
| Amplitudes a and b | No | The size of the waves and how deep the beats are |
| Time shown | No | Where the dots are on the traces |
| Given-answer mode | Sets ν2 | The page finds ν2 for a chosen upper answer, and warns if it falls outside 10 to 40 Hz |
| Play and time bar | No | Play moves the time from 0 to 0.5 s at 1/12 of real speed and holds 3 s at one beat of trial 1. The bar sets the time |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Frame frequencies with beat ν | ν1, ν2 | and Hz |
| Multiple of the beat | m | |
| Frequency of A in K′ | ν0 | Hz |
| Beat | ν | Hz |
| Amplitudes | a, b | and cm |
Exam traps
- Taking ν0 = 20 or 24 Hz. These are frequencies of the frame K′, not of A in K′.
- Subtracting 20 from 24 and calling the 4 Hz the beat. The beat is |ν̄ − ν0|, and ν0 sits half way between 20 and 24.
- Giving only one answer. There are two, one on each side of ν0.
- Writing the beat as the sum of the frequencies. It is the difference.