University Physics III · topic UPIII-01.02
The SHM equation of motion and its general solution
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.
The SHM equation and its general solution: predict a finite change
The SHM equation of motion is d²x/dt² = −ω²x, and its general solution is x(t) = A cos(ωt + φ): the equation fixes ω, while the amplitude A and the phase constant φ are fixed by the initial position and velocity. Use the supplied topic model Y_out = sqrt(x0^2+(v0/w)^2), where Y_out is AMPLITUDE A (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. w is Angular frequency ω, measured in rad s⁻¹. Keep x0 (Initial displacement x₀) = 0.04 m; v0 (Initial velocity v₀) = -0.6 m s⁻¹ fixed. For this exercise use only 7.0625 rad s⁻¹ ≤ w ≤ 14.9375 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. Calculate Y_out at w = 8.795 rad s⁻¹ and 9.8975 rad s⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
AMPLITUDE A against w
Predictions from the supplied The SHM equation and its general solution 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): AMPLITUDE A / m
Horizontal axis (x): w: Angular frequency ω / 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 ω / rad s⁻¹ | AMPLITUDE A / m |
|---|---|
| 7.0625 | 0.09390143680462458 |
| 9.03125 | 0.07754830918170535 |
| 11 | 0.06764027359177555 |
| 12.96875 | 0.06115926584827531 |
| 14.9375 | 0.05668701025855051 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 8.795 rad s⁻¹, substitute into Y_out = sqrt(x0^2+(v0/w)^2): Y₀ = 0.0790825 m.
- At w = 9.8975 rad s⁻¹, the same fixed-parameter model gives Y₁ = 0.0726289 m.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.00645368 m; the input change is Δw = 1.1025 rad s⁻¹.
- Average sensitivity is ΔY_out/Δw = -0.00585368 m per rad s⁻¹. Its negative sign means that this response decreases 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 = 0.0790825 m, Y₁ = 0.0726289 m; ΔY_out = -0.00645368 m; average sensitivity = -0.00585368 m per rad s⁻¹.
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.
The SHM equation and its general solution: infer an input with uncertainty
The SHM equation of motion is d²x/dt² = −ω²x, and its general solution is x(t) = A cos(ωt + φ): the equation fixes ω, while the amplitude A and the phase constant φ are fixed by the initial position and velocity. Use the supplied topic model Y_out = sqrt(x0^2+(v0/w)^2), where Y_out is AMPLITUDE A (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. w is Angular frequency ω, measured in rad s⁻¹. Keep x0 (Initial displacement x₀) = 0.04 m; v0 (Initial velocity v₀) = -0.6 m s⁻¹ fixed. For this exercise use only 7.0625 rad s⁻¹ ≤ w ≤ 14.9375 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 = 0.06447228 m, with a bounded reading uncertainty ±0.0011678 m. 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 0.0007875 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.
AMPLITUDE A against w
Predictions from the supplied The SHM equation and its general solution 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): AMPLITUDE A / m
Horizontal axis (x): w: Angular frequency ω / 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 ω / rad s⁻¹ | AMPLITUDE A / m |
|---|---|
| 7.0625 | 0.09390143680462458 |
| 9.03125 | 0.07754830918170535 |
| 11 | 0.06764027359177555 |
| 12.96875 | 0.06115926584827531 |
| 14.9375 | 0.05668701025855051 |
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(7.0625) = 0.0939014 m and Y_out(14.9375) = 0.056687 m. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 7.0625 + (0.0644723 − 0.0939014)(14.9375 − 7.0625)/(0.056687 − 0.0939014) = 13.29 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 = 11.8663 rad s⁻¹; substituting back gives Y_out = 0.06447228 m.
- The interpolation error in the input is |w_line − w| = 1.4238 rad s⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.06330448 m and 0.06564008 m. Sort the roots even if the response decreases: 11.5286 rad s⁻¹ ≤ w ≤ 12.2285 rad s⁻¹.
- The lower and upper input deviations are 0.337633 rad s⁻¹ and 0.362215 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 ≈ 11.8663 rad s⁻¹; compatible interval [11.5286, 12.2285] rad s⁻¹. Endpoint interpolation gives 13.29 rad s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.