University Physics III · unit UPIII-02
Mechanical Waves
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.
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.
Wave Function, Wave Equation & Wave Speed: infer an input with uncertainty
The linear wave equation states that the curvature of a disturbance in space equals its acceleration in time divided by v², so one constant v, fixed by the medium's stiffness and inertia, governs every wave the medium carries. Use the supplied topic model Y_out = ft/(0.001*mu), where Y_out is COEFFICIENT F_T/μ = v² (m² s⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. mu is Linear density μ, measured in g m⁻¹. Keep ft (Tension F_T) = 90 N; t (Elapsed time t) = 14 ms fixed. For this exercise use only 5.7 g m⁻¹ ≤ mu ≤ 18.3 g m⁻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 = 6723.4424 m² s⁻², with a bounded reading uncertainty ±270.81 m² s⁻². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown mu by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00126 g m⁻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.
COEFFICIENT F_T/μ = v² against mu
Predictions from the supplied Wave Function, Wave Equation & Wave Speed 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): COEFFICIENT F_T/μ = v² / m² s⁻²
Horizontal axis (x): mu: Linear density μ / g m⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| mu: Linear density μ / g m⁻¹ | COEFFICIENT F_T/μ = v² / m² s⁻² |
|---|---|
| 5.7 | 15789.473684210525 |
| 8.85 | 10169.49152542373 |
| 12 | 7500 |
| 15.149999999999999 | 5940.594059405941 |
| 18.299999999999997 | 4918.032786885246 |
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(mu) = 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(5.7) = 15789.5 m² s⁻² and Y_out(18.3) = 4918.03 m² s⁻². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives mu_line = 5.7 + (6723.44 − 15789.5)(18.3 − 5.7)/(4918.03 − 15789.5) = 16.2075 g m⁻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 mu = 13.386 g m⁻¹; substituting back gives Y_out = 6723.4424 m² s⁻².
- The interpolation error in the input is |mu_line − mu| = 2.82153 g m⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 6452.6324 m² s⁻² and 6994.2524 m² s⁻². Sort the roots even if the response decreases: 12.8677 g m⁻¹ ≤ mu ≤ 13.9478 g m⁻¹.
- The lower and upper input deviations are 0.518292 g m⁻¹ and 0.561796 g m⁻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 mu ≈ 13.386 g m⁻¹; compatible interval [12.8677, 13.9478] g m⁻¹. Endpoint interpolation gives 16.2075 g m⁻¹.
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
The average power of a mechanical wave is the energy it hands across a fixed point each second; its intensity is that power divided by the area of wavefront carrying it. Use the supplied topic model Y_out = 9.8696e-6*f^2*a^2, where Y_out is AVERAGE POWER (W). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f is Frequency f, measured in Hz. Keep a (Amplitude A) = 2.5 mm; t (Snapshot time t) = 4 ms fixed. For this exercise use only 66.375 Hz ≤ f ≤ 113.625 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 = 0.55902426 W, with a bounded reading uncertainty ±0.035605 W. 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 0.004725 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.
AVERAGE POWER 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): AVERAGE POWER / W
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 | AVERAGE POWER / W |
|---|---|
| 66.375 | 0.271761941953125 |
| 78.1875 | 0.3770979948632812 |
| 90 | 0.4996485 |
| 101.8125 | 0.6394134573632813 |
| 113.625 | 0.7963928669531251 |
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(66.375) = 0.271762 W and Y_out(113.625) = 0.796393 W. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives f_line = 66.375 + (0.559024 − 0.271762)(113.625 − 66.375)/(0.796393 − 0.271762) = 92.2468 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 = 95.1975 Hz; substituting back gives Y_out = 0.55902426 W.
- The interpolation error in the input is |f_line − f| = 2.9507 Hz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.52341926 W and 0.59462926 W. Sort the roots even if the response decreases: 92.116 Hz ≤ f ≤ 98.1823 Hz.
- The lower and upper input deviations are 3.0815 Hz and 2.98483 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 ≈ 95.1975 Hz; compatible interval [92.116, 98.1823] Hz. Endpoint interpolation gives 92.2468 Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Superposition, reflection, and interference: infer an input with uncertainty
Superposition says that where two waves overlap, the displacement of the medium is the algebraic sum of what each wave would produce alone, so the resultant is fixed by their phase difference and not by either wave separately. Use the supplied topic model Y_out = ((u-1)/(u+1))^2, where Y_out is POWER REFLECTED R (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. u is Speed ratio v₂ / v₁, dimensionless. Keep a (Incident height A) = 24 fixed. For this exercise use only 0.39125 ≤ u ≤ 0.95825 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 = 0.022901053, with a bounded reading uncertainty ±0.003367. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown u by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000567, 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.
POWER REFLECTED R against u
Predictions from the supplied Superposition, reflection, and interference 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): POWER REFLECTED R
Horizontal axis (x): u: Speed ratio v₂ / v₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| u: Speed ratio v₂ / v₁ | POWER REFLECTED R |
|---|---|
| 0.39125 | 0.19145538837345785 |
| 0.533 | 0.09280031522210434 |
| 0.67475 | 0.03771678251657251 |
| 0.8165 | 0.01020472563600824 |
| 0.95825 | 0.00045454479520295 |
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(u) = 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(0.39125) = 0.191455 and Y_out(0.95825) = 0.000454545. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives u_line = 0.39125 + (0.0229011 − 0.191455)(0.95825 − 0.39125)/(0.000454545 − 0.191455) = 0.891616. 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 u = 0.73712; substituting back gives Y_out = 0.022901053.
- The interpolation error in the input is |u_line − u| = 0.154496. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.019534053 and 0.026268053. Sort the roots even if the response decreases: 0.72106 ≤ u ≤ 0.754749.
- The lower and upper input deviations are 0.0160595 and 0.0176288. 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 u ≈ 0.73712; compatible interval [0.72106, 0.754749]. Endpoint interpolation gives 0.891616.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.