University Physics III · course UPIII
University Physics III
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.
Periodic Motion & the SHM Condition: infer an input with uncertainty
Simple harmonic motion is oscillation in which the acceleration is proportional to the displacement from equilibrium and directed opposite to it, a = −ω²x, with ω² one positive constant of the system. Use the supplied topic model Y_out = 2*pi/w, where Y_out is PERIOD T₀ (LINEAR) (s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Angular frequency ω₀, measured in rad s⁻¹. Keep n (Cubic term at 0.20 m ÷ linear term) = 0.15; a (Amplitude A) = 0.12 m fixed. For this exercise use only 3.6375 rad s⁻¹ ≤ w ≤ 8.3625 rad s⁻¹ 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 = 0.96371568 s, with a bounded reading uncertainty ±0.031854 s. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0004725 rad s⁻¹, 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.
PERIOD T₀ (LINEAR) against w
Predictions from the supplied Periodic Motion & the SHM Condition 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): PERIOD T₀ (LINEAR) / s
Horizontal axis (x): w: Angular frequency ω₀ / rad s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Angular frequency ω₀ / rad s⁻¹ | PERIOD T₀ (LINEAR) / s |
|---|---|
| 3.6375 | 1.727336166922223 |
| 4.8187500000000005 | 1.3039035656922615 |
| 6 | 1.0471975511965976 |
| 7.18125 | 0.874943123715173 |
| 8.3625 | 0.7513525031006978 |
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(w) = 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(3.6375) = 1.72734 s and Y_out(8.3625) = 0.751353 s. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 3.6375 + (0.963716 − 1.72734)(8.3625 − 3.6375)/(0.751353 − 1.72734) = 7.33439 rad s⁻¹. 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 w = 6.51975 rad s⁻¹; substituting back gives Y_out = 0.96371568 s.
- The interpolation error in the input is |w_line − w| = 0.814643 rad s⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.93186168 s and 0.99556968 s. Sort the roots even if the response decreases: 6.31115 rad s⁻¹ ≤ w ≤ 6.74262 rad s⁻¹.
- The lower and upper input deviations are 0.208604 rad s⁻¹ and 0.222866 rad s⁻¹. 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 w ≈ 6.51975 rad s⁻¹; compatible interval [6.31115, 6.74262] rad s⁻¹. Endpoint interpolation gives 7.33439 rad s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Describing travelling waves: infer an input with uncertainty
A travelling wave is a disturbance whose shape moves through a medium at a fixed speed, while each element of the medium only oscillates about its own equilibrium position and is left where it started. Use the supplied topic model Y_out = v/f, where Y_out is WAVELENGTH λ = v/f (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f is Frequency f (set by the source), measured in Hz. Keep v (Wave speed v (set by the medium)) = 12 m s⁻¹; d (Probe position x along the cord) = 0.6 m fixed. For this exercise use only 7.85 Hz ≤ f ≤ 14.15 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 = 1.026255 m, with a bounded reading uncertainty ±0.02673 m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 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.
WAVELENGTH λ = v/f against f
Predictions from the supplied Describing travelling waves 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): WAVELENGTH λ = v/f / m
Horizontal axis (x): f: Frequency f (set by the source) / Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Frequency f (set by the source) / Hz | WAVELENGTH λ = v/f / m |
|---|---|
| 7.85 | 1.5286624203821657 |
| 9.424999999999999 | 1.2732095490716182 |
| 11 | 1.0909090909090908 |
| 12.575 | 0.9542743538767396 |
| 14.149999999999999 | 0.8480565371024735 |
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(f) = 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(7.85) = 1.52866 m and Y_out(14.15) = 0.848057 m. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f_line = 7.85 + (1.02625 − 1.52866)(14.15 − 7.85)/(0.848057 − 1.52866) = 12.5005 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 f = 11.693 Hz; substituting back gives Y_out = 1.026255 m.
- The interpolation error in the input is |f_line − f| = 0.807513 Hz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.999525 m and 1.052985 m. Sort the roots even if the response decreases: 11.3962 Hz ≤ f ≤ 12.0057 Hz.
- The lower and upper input deviations are 0.296827 Hz and 0.312702 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 f ≈ 11.693 Hz; compatible interval [11.3962, 12.0057] Hz. Endpoint interpolation gives 12.5005 Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
Fermat's principle and the ray laws: infer an input with uncertainty
Fermat's principle states that between two points light follows a route whose optical path length ∫ n ds, and so its travel time, is stationary against small deformations of the route: unchanged to first order, not necessarily least. Use the supplied topic model Y_out = 33.356*(n1*sqrt(16+xc^2)+n2*sqrt(16+(6-xc)^2)), where Y_out is TRAVEL TIME T(x) (ps). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. xc is Crossing point x, measured in cm. Keep n1 (Index n₁ above) = 1; n2 (Index n₂ below) = 1.5 fixed. For this exercise use only 0.6375 cm ≤ xc ≤ 3.7875 cm 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 = 422.39157 ps, with a bounded reading uncertainty ±1.4901 ps. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown xc 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 cm, 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.
TRAVEL TIME T(x) against xc
Predictions from the supplied Fermat's principle and the ray laws 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): TRAVEL TIME T(x) / ps
Horizontal axis (x): xc: Crossing point x / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| xc: Crossing point x / cm | TRAVEL TIME T(x) / ps |
|---|---|
| 0.6375 | 469.8364902845865 |
| 1.4249999999999998 | 445.6973429551044 |
| 2.2124999999999995 | 428.093827792164 |
| 3 | 416.95000000000005 |
| 3.7874999999999996 | 412.45783945119405 |
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(xc) = 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(0.6375) = 469.836 ps and Y_out(3.7875) = 412.458 ps. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives xc_line = 0.6375 + (422.392 − 469.836)(3.7875 − 0.6375)/(412.458 − 469.836) = 3.24215 cm. 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 xc = 2.559 cm; substituting back gives Y_out = 422.39157 ps.
- The interpolation error in the input is |xc_line − xc| = 0.683154 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 420.90147 ps and 423.88167 ps. Sort the roots even if the response decreases: 2.46075 cm ≤ xc ≤ 2.66467 cm.
- The lower and upper input deviations are 0.0982527 cm and 0.105666 cm. 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 xc ≈ 2.559 cm; compatible interval [2.46075, 2.66467] cm. Endpoint interpolation gives 3.24215 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.