University Physics I · topic UPI-14.07
Damped 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.
Further practice
Work through a fresh case before tackling the extension.
Damped Oscillations: predict a finite change
A damped oscillator is a mass on a linear restoring force that also feels a drag proportional to its velocity, so its motion decays: as a shrinking sinusoid when the damping is light, and as a creep back to equilibrium with no oscillation when it is critical or heavier. Use the supplied topic model Y_out = sqrt(max(k/2-b*b/16,0)), where Y_out is RINGING ω′ (0 = NONE) (rad s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. b is Damping constant b, measured in kg s⁻¹. Keep k (Spring constant k) = 200 N m⁻¹ fixed. For this exercise use only 6.375 kg s⁻¹ ≤ b ≤ 39.45 kg 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 b = 13.6515 kg s⁻¹ and 18.282 kg s⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δb across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
RINGING ω′ (0 = NONE) against b
Predictions from the supplied Damped Oscillations 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): RINGING ω′ (0 = NONE) / rad s⁻¹
Horizontal axis (x): b: Damping constant b / kg s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Damping constant b / kg s⁻¹ | RINGING ω′ (0 = NONE) / rad s⁻¹ |
|---|---|
| 6.375 | 9.872181164134904 |
| 14.643749999999999 | 9.305779742777805 |
| 22.912499999999998 | 8.196864277537784 |
| 31.18125 | 6.26363337267945 |
| 39.449999999999996 | 1.6526021148479835 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of b separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At b = 13.6515 kg s⁻¹, substitute into Y_out = sqrt(max(k/2-b*b/16,0)): Y₀ = 9.39959 rad s⁻¹.
- At b = 18.282 kg s⁻¹, the same fixed-parameter model gives Y₁ = 8.89441 rad s⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.50518 rad s⁻¹; the input change is Δb = 4.6305 kg s⁻¹.
- Average sensitivity is ΔY_out/Δb = -0.109098 rad s⁻¹ per kg 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 = 9.39959 rad s⁻¹, Y₁ = 8.89441 rad s⁻¹; ΔY_out = -0.50518 rad s⁻¹; average sensitivity = -0.109098 rad s⁻¹ per kg 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.
Damped Oscillations: infer an input with uncertainty
A damped oscillator is a mass on a linear restoring force that also feels a drag proportional to its velocity, so its motion decays: as a shrinking sinusoid when the damping is light, and as a creep back to equilibrium with no oscillation when it is critical or heavier. Use the supplied topic model Y_out = sqrt(max(k/2-b*b/16,0)), where Y_out is RINGING ω′ (0 = NONE) (rad s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. b is Damping constant b, measured in kg s⁻¹. Keep k (Spring constant k) = 200 N m⁻¹ fixed. For this exercise use only 6.375 kg s⁻¹ ≤ b ≤ 39.45 kg 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 = 7.4793786 rad s⁻¹, with a bounded reading uncertainty ±0.35892 rad 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 b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0033075 kg 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.
RINGING ω′ (0 = NONE) against b
Predictions from the supplied Damped Oscillations 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): RINGING ω′ (0 = NONE) / rad s⁻¹
Horizontal axis (x): b: Damping constant b / kg s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Damping constant b / kg s⁻¹ | RINGING ω′ (0 = NONE) / rad s⁻¹ |
|---|---|
| 6.375 | 9.872181164134904 |
| 14.643749999999999 | 9.305779742777805 |
| 22.912499999999998 | 8.196864277537784 |
| 31.18125 | 6.26363337267945 |
| 39.449999999999996 | 1.6526021148479835 |
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(b) = 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(6.375) = 9.87218 rad s⁻¹ and Y_out(39.45) = 1.6526 rad s⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b_line = 6.375 + (7.47938 − 9.87218)(39.45 − 6.375)/(1.6526 − 9.87218) = 16.0035 kg 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 b = 26.5508 kg s⁻¹; substituting back gives Y_out = 7.4793786 rad s⁻¹.
- The interpolation error in the input is |b_line − b| = 10.5473 kg s⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 7.1204586 rad s⁻¹ and 7.8382986 rad s⁻¹. Sort the roots even if the response decreases: 24.839 kg s⁻¹ ≤ b ≤ 28.0853 kg s⁻¹.
- The lower and upper input deviations are 1.71172 kg s⁻¹ and 1.53457 kg 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 b ≈ 26.5508 kg s⁻¹; compatible interval [24.839, 28.0853] kg s⁻¹. Endpoint interpolation gives 16.0035 kg s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.