University Physics V · course UPV
University Physics V
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.
Complex amplitudes and the phase that survives: infer an input with uncertainty
A complex amplitude z = r e^(iφ) carries a modulus, which fixes the probability through |z|² = z*z = r², and a phase, which is measurable only as a difference — when two amplitudes for the same outcome are added before the modulus is squared. Use the supplied topic model Y_out = 2*r2/max(1 + r2*r2, 0.01), where Y_out is VISIBILITY (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. r2 is Modulus r₂ of z₂, dimensionless. Keep d (Relative phase δ) = 60 °; a (Global phase α) = 0 ° fixed. For this exercise use only 0.10625 ≤ r2 ≤ 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 r2 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 against r2
Predictions from the supplied Complex amplitudes and the phase that survives 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
Horizontal axis (x): r2: Modulus r₂ of z₂
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r2: Modulus r₂ of z₂ | VISIBILITY |
|---|---|
| 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(r2) = 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 r2_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 r2 = 0.586625; substituting back gives Y_out = 0.87287016.
- The interpolation error in the input is |r2_line − r2| = 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 ≤ r2 ≤ 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 r2 ≈ 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.
Four quantitative failures of the classical account: infer an input with uncertainty
The classical failure inventory is the list of sharp wrong predictions — atomic collapse in picoseconds, harmonic spectra, temperature-blind heat capacities, an undefined entropy constant — that classical mechanics, electrodynamics and statistical mechanics make about ordinary matter. Use the supplied topic model Y_out = 1.5+(tr/t)^2*exp(-tr/t)/max((1-exp(-tr/t))^2,0.000000001)+(tv/t)^2*exp(-tv/t)/max((1-exp(-tv/t))^2,0.000000001), where Y_out is C_V AT T (R). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. tv is Vibration quantum θ_vib, measured in K. Keep tr (Rotation quantum θ_rot) = 85 K; t (Gas temperature T) = 300 K fixed. For this exercise use only 1384.38 K ≤ tv ≤ 5715.63 K 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 = 2.4936041 R, with a bounded reading uncertainty ±0.000039783 R. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown tv by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.433125 K, 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.
C_V AT T against tv
Predictions from the supplied Four quantitative failures of the classical account 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): C_V AT T / R
Horizontal axis (x): tv: Vibration quantum θ_vib / K
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tv: Vibration quantum θ_vib / K | C_V AT T / R |
|---|---|
| 1384.375 | 2.7085280607810933 |
| 2467.1875 | 2.511482706259243 |
| 3550 | 2.49435336253227 |
| 4632.8125 | 2.4933838079367443 |
| 5715.625 | 2.4933388825594234 |
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(tv) = 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(1384.38) = 2.70853 R and Y_out(5715.63) = 2.49334 R. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives tv_line = 1384.38 + (2.4936 − 2.70853)(5715.63 − 1384.38)/(2.49334 − 2.70853) = 5710.29 K. 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 tv = 4026.42 K; substituting back gives Y_out = 2.4936041 R.
- The interpolation error in the input is |tv_line − tv| = 1683.86 K. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.4935643 R and 2.4936439 R. Sort the roots even if the response decreases: 3977.43 K ≤ tv ≤ 4083.2 K.
- The lower and upper input deviations are 48.9911 K and 56.775 K. 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 tv ≈ 4026.42 K; compatible interval [3977.43, 4083.2] K. Endpoint interpolation gives 5710.29 K.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The state vector and its position representation: infer an input with uncertainty
The state of the particle is a unit vector |ψ⟩ in the Hilbert space L²(ℝ), defined only up to a global phase; its position representation ψ(x) = ⟨x|ψ⟩ lists its components along the delta-normalised position basis. Use the supplied topic model Y_out = (2*s1*s2/(s1^2+s2^2))*exp(-d^2/(s1^2+s2^2)), where Y_out is PROBABILITY |⟨φ|ψ⟩|² (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. d is Separation d, dimensionless. Keep s1 (Width σ₁ of φ) = 1; s2 (Width σ₂ of ψ) = 1 fixed. For this exercise use only 0.425 ≤ d ≤ 3.575 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 = 0.063734381, with a bounded reading uncertainty ±0.0093085. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000315, 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.
PROBABILITY |⟨φ|ψ⟩|² against d
Predictions from the supplied The state vector and its position representation 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): PROBABILITY |⟨φ|ψ⟩|²
Horizontal axis (x): d: Separation d
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Separation d | PROBABILITY |⟨φ|ψ⟩|² |
|---|---|
| 0.425 | 0.913645626396729 |
| 1.2125 | 0.4794679990742081 |
| 2 | 0.1353352832366127 |
| 2.7874999999999996 | 0.020546223525013892 |
| 3.5749999999999997 | 0.0016777318282638824 |
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(d) = 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(0.425) = 0.913646 and Y_out(3.575) = 0.00167773. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.425 + (0.0637344 − 0.913646)(3.575 − 0.425)/(0.00167773 − 0.913646) = 3.36065. 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 d = 2.3465; substituting back gives Y_out = 0.063734381.
- The interpolation error in the input is |d_line − d| = 1.01415. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.054425881 and 0.073042881. Sort the roots even if the response decreases: 2.28767 ≤ d ≤ 2.41285.
- The lower and upper input deviations are 0.0588337 and 0.0663471. 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 d ≈ 2.3465; compatible interval [2.28767, 2.41285]. Endpoint interpolation gives 3.36065.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Hamiltonian as a self-adjoint operator: infer an input with uncertainty
The Hamiltonian is the self-adjoint energy operator built by substituting x̂ and p̂ = −iℏ d/dx into the classical energy, together with the domain of states on which its boundary terms vanish and D(Ĥ†) = D(Ĥ). Use the supplied topic model Y_out = b, where Y_out is ψ AT LEFT WALL β (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Wall leak β = ψ(0), dimensionless. Keep a (Interior amplitude α) = 1 fixed. For this exercise use only -0.4725 ≤ b ≤ 0.4725 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 = 0.10395, with a bounded reading uncertainty ±0.055282. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000945, 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.
ψ AT LEFT WALL β against b
Predictions from the supplied The Hamiltonian as a self-adjoint operator 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): ψ AT LEFT WALL β
Horizontal axis (x): b: Wall leak β = ψ(0)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Wall leak β = ψ(0) | ψ AT LEFT WALL β |
|---|---|
| -0.4725 | -0.4725 |
| -0.23625000000000002 | -0.23625000000000002 |
| -5.551115123125783e-17 | -5.551115123125783e-17 |
| 0.2362499999999999 | 0.2362499999999999 |
| 0.47249999999999986 | 0.47249999999999986 |
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(b) = 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(-0.4725) = -0.4725 and Y_out(0.4725) = 0.4725. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = -0.4725 + (0.10395 − -0.4725)(0.4725 − -0.4725)/(0.4725 − -0.4725) = 0.10395. 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 b = 0.10395; substituting back gives Y_out = 0.10395.
- The interpolation error in the input is |b_line − b| = 5.55112e-17. This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.048668 and 0.159232. Sort the roots even if the response decreases: 0.048668 ≤ b ≤ 0.159232.
- The lower and upper input deviations are 0.055282 and 0.055282. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer b ≈ 0.10395; compatible interval [0.048668, 0.159232]. Endpoint interpolation gives 0.10395.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.