University Physics III · topic UPIII-02.05
Standing waves and normal modes
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.
Further practice
Work through a fresh case before tackling the extension.
Standing Waves & Normal Modes: predict a finite change
A standing wave is the motion a bounded medium settles into when two identical waves travel through it in opposite directions: every point oscillates at one frequency with an amplitude fixed by its position, and the nodes stay put. Use the supplied topic model Y_out = amp*cosd(ph), where Y_out is ANTINODE NOW (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. ph is Phase ωt, measured in °. Keep q (Quarter wavelengths fitted into L) = 4; amp (Antinode amplitude) = 8 mm fixed. For this exercise use only 189.45 ° ≤ ph ≤ 321.75 ° 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. Calculate Y_out at ph = 218.556 ° and 237.078 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δph across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ANTINODE NOW against ph
Predictions from the supplied Standing Waves & Normal Modes 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): ANTINODE NOW / mm
Horizontal axis (x): ph: Phase ωt / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ph: Phase ωt / ° | ANTINODE NOW / mm |
|---|---|
| 189.45 | -7.891434057411964 |
| 222.52499999999998 | -5.895859878383609 |
| 255.6 | -1.9895190973188424 |
| 288.675 | 2.5615972930184974 |
| 321.75 | 6.282535447045957 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of ph separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At ph = 218.556 °, substitute into Y_out = amp*cosd(ph): Y₀ = -6.25599 mm.
- At ph = 237.078 °, the same fixed-parameter model gives Y₁ = -4.34797 mm.
- Signed change: ΔY_out = Y₁ − Y₀ = 1.90802 mm; the input change is Δph = 18.522 °.
- Average sensitivity is ΔY_out/Δph = 0.103014 mm per °. Its positive sign means that this response increases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = -6.25599 mm, Y₁ = -4.34797 mm; ΔY_out = 1.90802 mm; average sensitivity = 0.103014 mm per °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Standing Waves & Normal Modes: infer an input with uncertainty
A standing wave is the motion a bounded medium settles into when two identical waves travel through it in opposite directions: every point oscillates at one frequency with an amplitude fixed by its position, and the nodes stay put. Use the supplied topic model Y_out = amp*cosd(ph), where Y_out is ANTINODE NOW (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. ph is Phase ωt, measured in °. Keep q (Quarter wavelengths fitted into L) = 4; amp (Antinode amplitude) = 8 mm fixed. For this exercise use only 189.45 ° ≤ ph ≤ 321.75 ° 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 = 0.021362805 mm, with a bounded reading uncertainty ±0.93918 mm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown ph by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.01323 °, 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.
ANTINODE NOW against ph
Predictions from the supplied Standing Waves & Normal Modes 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): ANTINODE NOW / mm
Horizontal axis (x): ph: Phase ωt / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ph: Phase ωt / ° | ANTINODE NOW / mm |
|---|---|
| 189.45 | -7.891434057411964 |
| 222.52499999999998 | -5.895859878383609 |
| 255.6 | -1.9895190973188424 |
| 288.675 | 2.5615972930184974 |
| 321.75 | 6.282535447045957 |
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(ph) = 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(189.45) = -7.89143 mm and Y_out(321.75) = 6.28254 mm. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives ph_line = 189.45 + (0.0213628 − -7.89143)(321.75 − 189.45)/(6.28254 − -7.89143) = 263.308 °. 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 ph = 270.153 °; substituting back gives Y_out = 0.021362805 mm.
- The interpolation error in the input is |ph_line − ph| = 6.84486 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.9178172 mm and 0.9605428 mm. Sort the roots even if the response decreases: 263.412 ° ≤ ph ≤ 276.896 °.
- The lower and upper input deviations are 6.74089 ° and 6.74302 °. 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 ph ≈ 270.153 °; compatible interval [263.412, 276.896] °. Endpoint interpolation gives 263.308 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.