University Physics I · unit UPI-EXT-WS
Institutional Extension · Waves & Sound
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.
Mechanical Waves & Representations: infer an input with uncertainty
A mechanical wave is a disturbance that travels through a material medium at a speed set by that medium, carrying energy from place to place while each particle of the medium only oscillates about its own equilibrium position. Use the supplied topic model Y_out = (p-2)/v, where Y_out is PULSE REACHES PROBE AT t (s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v is Wave speed v, measured in m/s. Keep t (Time now t) = 5 s; p (Probe position x₀) = 6 m fixed. For this exercise use only 0.9275 m/s ≤ v ≤ 1.8725 m/s 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 = 2.6596629 s, with a bounded reading uncertainty ±0.078522 s. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v 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 m/s, 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 REACHES PROBE AT t against v
Predictions from the supplied Mechanical Waves & Representations 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 REACHES PROBE AT t / s
Horizontal axis (x): v: Wave speed v / m/s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Wave speed v / m/s | PULSE REACHES PROBE AT t / s |
|---|---|
| 0.9275 | 4.3126684636118595 |
| 1.1637499999999998 | 3.437164339419979 |
| 1.4 | 2.857142857142857 |
| 1.63625 | 2.4446142093200915 |
| 1.8724999999999998 | 2.136181575433912 |
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(v) = 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(0.9275) = 4.31267 s and Y_out(1.8725) = 2.13618 s. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 0.9275 + (2.65966 − 4.31267)(1.8725 − 0.9275)/(2.13618 − 4.31267) = 1.64521 m/s. 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 v = 1.50395 m/s; substituting back gives Y_out = 2.6596629 s.
- The interpolation error in the input is |v_line − v| = 0.141262 m/s. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.5811409 s and 2.7381849 s. Sort the roots even if the response decreases: 1.46082 m/s ≤ v ≤ 1.5497 m/s.
- The lower and upper input deviations are 0.0431283 m/s and 0.0457523 m/s. 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 v ≈ 1.50395 m/s; compatible interval [1.46082, 1.5497] m/s. Endpoint interpolation gives 1.64521 m/s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Sinusoidal Wave Functions & Phase: infer an input with uncertainty
A sinusoidal travelling wave is a disturbance y(x, t) = A sin(kx − ωt + φ₀) that depends on position and time only through the phase kx − ωt + φ₀, so the same sine pattern repeats every λ = 2π/k in space and every T = 2π/ω in time. Use the supplied topic model Y_out = 2*pi/lam, where Y_out is WAVE NUMBER k = 2π/λ (rad m⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lam is Wavelength λ, measured in m. Keep t (Time t) = 0.4 s; dir (Direction of travel (−1 or +1)) = 1 fixed. For this exercise use only 1.31875 m ≤ lam ≤ 3.68125 m 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 = 2.2766195 rad m⁻¹, with a bounded reading uncertainty ±0.085472 rad m⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lam 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 m, 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.
WAVE NUMBER k = 2π/λ against lam
Predictions from the supplied Sinusoidal Wave Functions & Phase 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): WAVE NUMBER k = 2π/λ / rad m⁻¹
Horizontal axis (x): lam: Wavelength λ / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam: Wavelength λ / m | WAVE NUMBER k = 2π/λ / rad m⁻¹ |
|---|---|
| 1.31875 | 4.764500706866037 |
| 1.909375 | 3.290702615871469 |
| 2.5 | 2.5132741228718345 |
| 3.090625 | 2.0329821014130105 |
| 3.68125 | 1.7068075537329945 |
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(lam) = 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(1.31875) = 4.7645 rad m⁻¹ and Y_out(3.68125) = 1.70681 rad m⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lam_line = 1.31875 + (2.27662 − 4.7645)(3.68125 − 1.31875)/(1.70681 − 4.7645) = 3.24099 m. 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 lam = 2.75988 m; substituting back gives Y_out = 2.2766195 rad m⁻¹.
- The interpolation error in the input is |lam_line − lam| = 0.481115 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.1911475 rad m⁻¹ and 2.3620915 rad m⁻¹. Sort the roots even if the response decreases: 2.66001 m ≤ lam ≤ 2.86753 m.
- The lower and upper input deviations are 0.0998657 m and 0.107657 m. 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 lam ≈ 2.75988 m; compatible interval [2.66001, 2.86753] m. Endpoint interpolation gives 3.24099 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Wave Speed in Strings & Media: infer an input with uncertainty
Wave speed is the rate at which a disturbance travels through a medium, fixed by the medium's stiffness and inertia — on a stretched string v = √(F_T/μ) — and tied to the source frequency by the identity v = fλ. Use the supplied topic model Y_out = sqrt(1000*Ft/mu)/f, where Y_out is WAVELENGTH λ (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. f is Driver frequency f, measured in Hz. Keep Ft (Tension F_T) = 72 N; mu (Linear density μ) = 6.3 g m⁻¹ fixed. For this exercise use only 97 Hz ≤ f ≤ 223 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 = 0.6148884 m, with a bounded reading uncertainty ±0.020324 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 0.0126 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 against f
Predictions from the supplied Wave Speed in Strings & 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): WAVELENGTH λ / m
Horizontal axis (x): f: Driver frequency f / Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Driver frequency f / Hz | WAVELENGTH λ / m |
|---|---|
| 97 | 1.1021082140718532 |
| 128.5 | 0.8319416090659125 |
| 160 | 0.668153104781061 |
| 191.5 | 0.5582480248823486 |
| 223 | 0.4793923621747523 |
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(97) = 1.10211 m and Y_out(223) = 0.479392 m. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f_line = 97 + (0.614888 − 1.10211)(223 − 97)/(0.479392 − 1.10211) = 195.584 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 = 173.86 Hz; substituting back gives Y_out = 0.6148884 m.
- The interpolation error in the input is |f_line − f| = 21.7238 Hz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.5945644 m and 0.6352124 m. Sort the roots even if the response decreases: 168.297 Hz ≤ f ≤ 179.803 Hz.
- The lower and upper input deviations are 5.56275 Hz and 5.94306 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 ≈ 173.86 Hz; compatible interval [168.297, 179.803] Hz. Endpoint interpolation gives 195.584 Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Wave Energy, Power & Intensity: infer an input with uncertainty
A travelling wave transports energy without transporting matter. On a string its average power is P = ½μvω²A², and its intensity is that power divided by the wavefront area it crosses, measured in W m⁻². Use the supplied topic model Y_out = sqrt(20*ft)/f, where Y_out is WAVELENGTH λ (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. f is Frequency f, measured in Hz. Keep a (Amplitude A) = 5 mm; ft (Tension F) = 80 N fixed. For this exercise use only 43.1875 Hz ≤ f ≤ 66.8125 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 = 0.69445952 m, with a bounded reading uncertainty ±0.014365 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 0.0023625 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 against f
Predictions from the supplied Wave Energy, Power & Intensity 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 / m
Horizontal axis (x): f: Frequency f / Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Frequency f / Hz | WAVELENGTH λ / m |
|---|---|
| 43.1875 | 0.9261939218523878 |
| 49.09375 | 0.8147676639083387 |
| 55 | 0.7272727272727273 |
| 60.90625 | 0.6567470497691124 |
| 66.8125 | 0.598690364826941 |
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(43.1875) = 0.926194 m and Y_out(66.8125) = 0.59869 m. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f_line = 43.1875 + (0.69446 − 0.926194)(66.8125 − 43.1875)/(0.59869 − 0.926194) = 59.904 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 = 57.5987 Hz; substituting back gives Y_out = 0.69445952 m.
- The interpolation error in the input is |f_line − f| = 2.30529 Hz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.68009452 m and 0.70882452 m. Sort the roots even if the response decreases: 56.4315 Hz ≤ f ≤ 58.8154 Hz.
- The lower and upper input deviations are 1.16729 Hz and 1.2166 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 ≈ 57.5987 Hz; compatible interval [56.4315, 58.8154] Hz. Endpoint interpolation gives 59.904 Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.