University Physics III · unit UPIII-03
Wave Phenomena
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Superposition & Linear Media: infer an input with uncertainty
The superposition principle says that where two or more waves overlap in a linear medium, the displacement of the medium at each point and instant is the algebraic sum of the displacements each wave would produce there on its own. Use the supplied topic model Y_out = 15*exp(-((110-2.75*t)/26)^2), where Y_out is PULSE 1 AT MEETING POINT (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. t is Time t, measured in ms. Keep b (Second pulse height A₂) = -10 mm; eps (Nonlinear mixing ε) = 0 fixed. For this exercise use only 6.375 ms ≤ t ≤ 37.875 ms 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 = 1.4697082 mm, with a bounded reading uncertainty ±0.22045 mm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00315 ms, 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.
PULSE 1 AT MEETING POINT against t
Predictions from the supplied Superposition & Linear Media 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): PULSE 1 AT MEETING POINT / mm
Horizontal axis (x): t: Time t / ms
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time t / ms | PULSE 1 AT MEETING POINT / mm |
|---|---|
| 6.375 | 0.00004817965821316306 |
| 14.25 | 0.009007332580196772 |
| 22.125 | 0.42045715745713724 |
| 30 | 4.900499718827123 |
| 37.875 | 14.26106813408309 |
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(t) = 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(6.375) = 0.0000481797 mm and Y_out(37.875) = 14.2611 mm. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 6.375 + (1.46971 − 0.0000481797)(37.875 − 6.375)/(14.2611 − 0.0000481797) = 9.62121 ms. 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 t = 25.59 ms; substituting back gives Y_out = 1.4697082 mm.
- The interpolation error in the input is |t_line − t| = 15.9688 ms. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.2492582 mm and 1.6901582 mm. Sort the roots even if the response decreases: 25.0945 ms ≤ t ≤ 26.0302 ms.
- The lower and upper input deviations are 0.495535 ms and 0.440199 ms. 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 t ≈ 25.59 ms; compatible interval [25.0945, 26.0302] ms. Endpoint interpolation gives 9.62121 ms.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Phase difference, path difference, and coherence: infer an input with uncertainty
The phase difference is the single number δ saying how far one arriving wave's cycle sits behind another's at the same point, built from the path difference and from whatever offset the sources already carried. Use the supplied topic model Y_out = exp(-sig*sig/2), where Y_out is COHERENCE FACTOR (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. sig is Phase jitter σ, measured in rad. Keep dr (Path difference Δr / λ) = 0.35 λ; p0 (Source phase offset Δφ₀) = 0.5 π rad fixed. For this exercise use only 0.31875 rad ≤ sig ≤ 2.68125 rad 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 = 0.21254958, with a bounded reading uncertainty ±0.027761. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown sig by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625 rad, 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.
COHERENCE FACTOR against sig
Predictions from the supplied Phase difference, path difference, and coherence 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): COHERENCE FACTOR
Horizontal axis (x): sig: Phase jitter σ / rad
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sig: Phase jitter σ / rad | COHERENCE FACTOR |
|---|---|
| 0.31875 | 0.9504680027199568 |
| 0.9093749999999999 | 0.6613438052942324 |
| 1.5 | 0.32465246735834974 |
| 2.0906249999999997 | 0.11243764863785047 |
| 2.68125 | 0.027473021800949925 |
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(sig) = 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(0.31875) = 0.950468 and Y_out(2.68125) = 0.027473. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives sig_line = 0.31875 + (0.21255 − 0.950468)(2.68125 − 0.31875)/(0.027473 − 0.950468) = 2.20753 rad. 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 sig = 1.75987 rad; substituting back gives Y_out = 0.21254958.
- The interpolation error in the input is |sig_line − sig| = 0.447653 rad. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.18478858 and 0.24031058. Sort the roots even if the response decreases: 1.68868 rad ≤ sig ≤ 1.83768 rad.
- The lower and upper input deviations are 0.0711932 rad and 0.0778099 rad. 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 sig ≈ 1.75987 rad; compatible interval [1.68868, 1.83768] rad. Endpoint interpolation gives 2.20753 rad.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Amplitude, Intensity & Fringe Visibility: infer an input with uncertainty
The resultant intensity of two overlapping coherent waves is the sum of their separate intensities plus an interference term, 2√(I₁I₂) cos δ, controlled entirely by the phase difference δ between them. Use the supplied topic model Y_out = 2*r*gam/(1+r^2), where Y_out is VISIBILITY V (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Amplitude ratio A₂/A₁, dimensionless. Keep gam (Degree of coherence |γ|) = 1 fixed. For this exercise use only 0.10625 ≤ r ≤ 0.89375 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 = 0.87287016, with a bounded reading uncertainty ±0.018128. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, 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.
VISIBILITY V against r
Predictions from the supplied Amplitude, Intensity & Fringe Visibility 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): VISIBILITY V
Horizontal axis (x): r: Amplitude ratio A₂/A₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Amplitude ratio A₂/A₁ | VISIBILITY V |
|---|---|
| 0.10625 | 0.21012785352852564 |
| 0.303125 | 0.5552325841390228 |
| 0.5 | 0.8 |
| 0.6968749999999999 | 0.93815117433231 |
| 0.8937499999999999 | 0.9937240765271774 |
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(r) = 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(0.10625) = 0.210128 and Y_out(0.89375) = 0.993724. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives r_line = 0.10625 + (0.87287 − 0.210128)(0.89375 − 0.10625)/(0.993724 − 0.210128) = 0.772294. 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 r = 0.586625; substituting back gives Y_out = 0.87287016.
- The interpolation error in the input is |r_line − r| = 0.185669. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.85474216 and 0.89099816. Sort the roots even if the response decreases: 0.562681 ≤ r ≤ 0.612788.
- The lower and upper input deviations are 0.023944 and 0.0261631. 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 r ≈ 0.586625; compatible interval [0.562681, 0.612788]. Endpoint interpolation gives 0.772294.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Nodal Lines & Accessible Orders: infer an input with uncertainty
A nodal line is the locus of points whose distances from two coherent sources differ by an odd number of half wavelengths — one branch of a hyperbola whose foci are the two sources. Use the supplied topic model Y_out = m/dl, where Y_out is SIN θ (ABOVE 1: NO BRANCH) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dl is Source separation d / λ, measured in λ. Keep m (Branch order m) = 1.5 fixed. For this exercise use only 0.871875 λ ≤ dl ≤ 3.62812 λ 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 = 0.58750092, with a bounded reading uncertainty ±0.02611. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dl by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000275625 λ, 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.
SIN θ (ABOVE 1: NO BRANCH) against dl
Predictions from the supplied Nodal Lines & Accessible Orders 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): SIN θ (ABOVE 1: NO BRANCH)
Horizontal axis (x): dl: Source separation d / λ / λ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dl: Source separation d / λ / λ | SIN θ (ABOVE 1: NO BRANCH) |
|---|---|
| 0.871875 | 1.7204301075268817 |
| 1.5609374999999999 | 0.9609609609609611 |
| 2.25 | 0.6666666666666666 |
| 2.9390624999999995 | 0.5103668261562999 |
| 3.628125 | 0.41343669250646 |
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(dl) = 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(0.871875) = 1.72043 and Y_out(3.62812) = 0.413437. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives dl_line = 0.871875 + (0.587501 − 1.72043)(3.62812 − 0.871875)/(0.413437 − 1.72043) = 3.26105 λ. 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 dl = 2.55319 λ; substituting back gives Y_out = 0.58750092.
- The interpolation error in the input is |dl_line − dl| = 0.707863 λ. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.56139092 and 0.61361092. Sort the roots even if the response decreases: 2.44455 λ ≤ dl ≤ 2.67193 λ.
- The lower and upper input deviations are 0.108642 λ and 0.118747 λ. 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 dl ≈ 2.55319 λ; compatible interval [2.44455, 2.67193] λ. Endpoint interpolation gives 3.26105 λ.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.