University Physics IV · topic UPIV-08.01
The classical oscillator and the harmonic approximation
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 classical oscillator and the harmonic approximation: predict a finite change
The harmonic approximation replaces a potential near a stable minimum by the leading quadratic term of its Taylor series, ½U″(x₀)ξ², so the system becomes a mass on a spring whose stiffness is the curvature of the curve. Use the supplied topic model Y_out = 1/(pi*sqrt(2*en/k)*sqrt(1-s*s)), where Y_out is DENSITY AT PROBE (/m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. s is Probe position x/A, measured in of A. Keep en (Total energy E) = 0.6 J; k (Curvature k = U″₀) = 3 N/m fixed. For this exercise use only -0.748125 of A ≤ s ≤ -0.049875 of A 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 s = -0.59451 of A and -0.496755 of A. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δs across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DENSITY AT PROBE against s
Predictions from the supplied The classical oscillator and the harmonic approximation 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): DENSITY AT PROBE / /m
Horizontal axis (x): s: Probe position x/A / of A
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Probe position x/A / of A | DENSITY AT PROBE / /m |
|---|---|
| -0.7481249999999999 | 0.7584751526560197 |
| -0.5735625 | 0.6143989130588873 |
| -0.39899999999999997 | 0.5488757434899616 |
| -0.22443749999999996 | 0.516468001905529 |
| -0.049875 | 0.5039192648176797 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of s separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At s = -0.59451 of A, substitute into Y_out = 1/(pi*sqrt(2*en/k)*sqrt(1-s*s)): Y₀ = 0.625917 /m.
- At s = -0.496755 of A, the same fixed-parameter model gives Y₁ = 0.579903 /m.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.046014 /m; the input change is Δs = 0.097755 of A.
- Average sensitivity is ΔY_out/Δs = -0.470707 /m per of A. 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 = 0.625917 /m, Y₁ = 0.579903 /m; ΔY_out = -0.046014 /m; average sensitivity = -0.470707 /m per of A.
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 classical oscillator and the harmonic approximation: infer an input with uncertainty
The harmonic approximation replaces a potential near a stable minimum by the leading quadratic term of its Taylor series, ½U″(x₀)ξ², so the system becomes a mass on a spring whose stiffness is the curvature of the curve. Use the supplied topic model Y_out = 1/(pi*sqrt(2*en/k)*sqrt(1-s*s)), where Y_out is DENSITY AT PROBE (/m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. s is Probe position x/A, measured in of A. Keep en (Total energy E) = 0.6 J; k (Curvature k = U″₀) = 3 N/m fixed. For this exercise use only -0.748125 of A ≤ s ≤ -0.049875 of A 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 = 0.53164239 /m, with a bounded reading uncertainty ±0.0041585 /m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000069825 of A, 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.
DENSITY AT PROBE against s
Predictions from the supplied The classical oscillator and the harmonic approximation 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): DENSITY AT PROBE / /m
Horizontal axis (x): s: Probe position x/A / of A
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Probe position x/A / of A | DENSITY AT PROBE / /m |
|---|---|
| -0.7481249999999999 | 0.7584751526560197 |
| -0.5735625 | 0.6143989130588873 |
| -0.39899999999999997 | 0.5488757434899616 |
| -0.22443749999999996 | 0.516468001905529 |
| -0.049875 | 0.5039192648176797 |
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(s) = 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(-0.748125) = 0.758475 /m and Y_out(-0.049875) = 0.503919 /m. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = -0.748125 + (0.531642 − 0.758475)(-0.049875 − -0.748125)/(0.503919 − 0.758475) = -0.12592 of A. 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 s = -0.322192 of A; substituting back gives Y_out = 0.53164239 /m.
- The interpolation error in the input is |s_line − s| = 0.196273 of A. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.52748389 /m and 0.53580089 /m. Sort the roots even if the response decreases: -0.343024 of A ≤ s ≤ -0.299369 of A.
- The lower and upper input deviations are 0.0208311 of A and 0.0228235 of A. 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 s ≈ -0.322192 of A; compatible interval [-0.343024, -0.299369] of A. Endpoint interpolation gives -0.12592 of A.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.