University Physics III · topic UPIII-01.01
Periodic motion and the simple-harmonic condition
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.
Periodic Motion & the SHM Condition: predict a finite change
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. Calculate Y_out at w = 4.677 rad s⁻¹ and 5.3385 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.
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
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 4.677 rad s⁻¹, substitute into Y_out = 2*pi/w: Y₀ = 1.34342 s.
- At w = 5.3385 rad s⁻¹, the same fixed-parameter model gives Y₁ = 1.17696 s.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.166465 s; the input change is Δw = 0.6615 rad s⁻¹.
- Average sensitivity is ΔY_out/Δw = -0.251648 s 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 = 1.34342 s, Y₁ = 1.17696 s; ΔY_out = -0.166465 s; average sensitivity = -0.251648 s 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.
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.