University Physics III · unit UPIII-01
Oscillations
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.
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.
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.
Velocity, acceleration and phase in SHM: infer an input with uncertainty
In simple harmonic motion the velocity and acceleration are the first and second time derivatives of the sinusoidal displacement, so each has an amplitude ω times the last and runs a quarter cycle ahead of it in phase. Use the supplied topic model Y_out = -w*w*amp*cos(w*tc), where Y_out is a AT CURSOR (m s⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. w is Angular frequency ω, measured in rad s⁻¹. Keep amp (Amplitude A) = 0.12 m; tc (Cursor time t) = 0.3 s fixed. For this exercise use only 3.74 rad s⁻¹ ≤ w ≤ 7.3625 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 = 0.90263253 m s⁻², with a bounded reading uncertainty ±0.24464 m 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 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 0.00036225 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.
a AT CURSOR against w
Predictions from the supplied Velocity, acceleration and phase in SHM 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): a AT CURSOR / m s⁻²
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 | a AT CURSOR / m s⁻² |
|---|---|
| 3.7399999999999998 | -0.7282750948965174 |
| 4.645625 | -0.45628576057990855 |
| 5.55125 | 0.34922744990951354 |
| 6.456875 | 1.7917146865531832 |
| 7.3625 | 3.8739335162845174 |
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(3.74) = -0.728275 m s⁻² and Y_out(7.3625) = 3.87393 m s⁻². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives w_line = 3.74 + (0.902633 − -0.728275)(7.3625 − 3.74)/(3.87393 − -0.728275) = 5.02372 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 = 5.94973 rad s⁻¹; substituting back gives Y_out = 0.90263253 m s⁻².
- The interpolation error in the input is |w_line − w| = 0.926002 rad s⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.65799253 m s⁻² and 1.1472725 m s⁻². Sort the roots even if the response decreases: 5.78467 rad s⁻¹ ≤ w ≤ 6.10167 rad s⁻¹.
- The lower and upper input deviations are 0.165051 rad s⁻¹ and 0.151947 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 ≈ 5.94973 rad s⁻¹; compatible interval [5.78467, 6.10167] rad s⁻¹. Endpoint interpolation gives 5.02372 rad s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Energy in simple harmonic motion: infer an input with uncertainty
The total mechanical energy of a simple harmonic oscillator is the constant sum E = ½mv² + ½mω²x², equal to ½mω²A², which the motion keeps converting between kinetic and potential form without changing the total. Use the supplied topic model Y_out = w*a*abs(sind(theta)), where Y_out is SPEED |v| (m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; sind/cosd/tand use degrees. theta is Phase ωt + φ, measured in °. Keep a (Amplitude A) = 0.2 m; w (Angular frequency ω) = 8 rad s⁻¹ fixed. For this exercise use only 94.95 ° ≤ theta ≤ 170.55 ° 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 = 1.0054796 m s⁻¹, with a bounded reading uncertainty ±0.088283 m s⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00756 °, 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.
SPEED |v| against theta
Predictions from the supplied Energy in simple harmonic motion 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): SPEED |v| / m s⁻¹
Horizontal axis (x): theta: Phase ωt + φ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Phase ωt + φ / ° | SPEED |v| / m s⁻¹ |
|---|---|
| 94.94999999999999 | 1.5940326023887061 |
| 113.85 | 1.463371455677782 |
| 132.75 | 1.1749160150970972 |
| 151.65 | 0.7597702237929513 |
| 170.55 | 0.2626989545101801 |
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(theta) = 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(94.95) = 1.59403 m s⁻¹ and Y_out(170.55) = 0.262699 m s⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives theta_line = 94.95 + (1.00548 − 1.59403)(170.55 − 94.95)/(0.262699 − 1.59403) = 128.371 °. 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 theta = 141.066 °; substituting back gives Y_out = 1.0054796 m s⁻¹.
- The interpolation error in the input is |theta_line − theta| = 12.6949 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.9171966 m s⁻¹ and 1.0937626 m s⁻¹. Sort the roots even if the response decreases: 136.874 ° ≤ theta ≤ 145.023 °.
- The lower and upper input deviations are 4.19173 ° and 3.95698 °. 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 theta ≈ 141.066 °; compatible interval [136.874, 145.023] °. Endpoint interpolation gives 128.371 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.