University Physics III · topic UPIII-03.05
Beats as amplitude modulation in time
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Beats as amplitude modulation: predict a finite change
A beat is the slow rise and fall in loudness heard when two waves of nearly equal frequency overlap, repeating at a rate equal to the difference between the two frequencies. Use the supplied topic model Y_out = 1/max(df,0.5), where Y_out is BEAT PERIOD 1/Δf (s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. df is Frequency difference Δf, measured in Hz. Keep f0 (Carrier (mean) frequency f₀) = 14 Hz; r (Amplitude ratio A₂/A₁) = 1 fixed. For this exercise use only 1.74375 Hz ≤ df ≤ 7.25625 Hz on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at df = 2.9565 Hz and 3.72825 Hz. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δdf across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
BEAT PERIOD 1/Δf against df
Predictions from the supplied Beats as amplitude modulation model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): BEAT PERIOD 1/Δf / s
Horizontal axis (x): df: Frequency difference Δf / Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| df: Frequency difference Δf / Hz | BEAT PERIOD 1/Δf / s |
|---|---|
| 1.74375 | 0.5734767025089607 |
| 3.1218749999999997 | 0.3203203203203204 |
| 4.5 | 0.2222222222222222 |
| 5.878124999999999 | 0.1701222753854333 |
| 7.25625 | 0.13781223083548666 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of df separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At df = 2.9565 Hz, substitute into Y_out = 1/max(df,0.5): Y₀ = 0.338238 s.
- At df = 3.72825 Hz, the same fixed-parameter model gives Y₁ = 0.268222 s.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0700154 s; the input change is Δdf = 0.77175 Hz.
- Average sensitivity is ΔY_out/Δdf = -0.0907229 s per Hz. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.338238 s, Y₁ = 0.268222 s; ΔY_out = -0.0700154 s; average sensitivity = -0.0907229 s per Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Beats as amplitude modulation: infer an input with uncertainty
A beat is the slow rise and fall in loudness heard when two waves of nearly equal frequency overlap, repeating at a rate equal to the difference between the two frequencies. Use the supplied topic model Y_out = 1/max(df,0.5), where Y_out is BEAT PERIOD 1/Δf (s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. df is Frequency difference Δf, measured in Hz. Keep f0 (Carrier (mean) frequency f₀) = 14 Hz; r (Amplitude ratio A₂/A₁) = 1 fixed. For this exercise use only 1.74375 Hz ≤ df ≤ 7.25625 Hz on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.19583364 s, with a bounded reading uncertainty ±0.0087032 s. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown df by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00055125 Hz, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
BEAT PERIOD 1/Δf against df
Predictions from the supplied Beats as amplitude modulation model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): BEAT PERIOD 1/Δf / s
Horizontal axis (x): df: Frequency difference Δf / Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| df: Frequency difference Δf / Hz | BEAT PERIOD 1/Δf / s |
|---|---|
| 1.74375 | 0.5734767025089607 |
| 3.1218749999999997 | 0.3203203203203204 |
| 4.5 | 0.2222222222222222 |
| 5.878124999999999 | 0.1701222753854333 |
| 7.25625 | 0.13781223083548666 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(df) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.74375) = 0.573477 s and Y_out(7.25625) = 0.137812 s. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives df_line = 1.74375 + (0.195834 − 0.573477)(7.25625 − 1.74375)/(0.137812 − 0.573477) = 6.5221 Hz. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives df = 5.10637 Hz; substituting back gives Y_out = 0.19583364 s.
- The interpolation error in the input is |df_line − df| = 1.41573 Hz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.18713044 s and 0.20453684 s. Sort the roots even if the response decreases: 4.88909 Hz ≤ df ≤ 5.34387 Hz.
- The lower and upper input deviations are 0.21728 Hz and 0.237491 Hz. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer df ≈ 5.10637 Hz; compatible interval [4.88909, 5.34387] Hz. Endpoint interpolation gives 6.5221 Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.