University Physics I · topic UPI-14.05
Mass–spring systems
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.
Mass–Spring Systems: predict a finite change
A mass–spring system is a body of mass m held by one or more springs whose net force about equilibrium is −k_eff times the displacement, so it oscillates with ω = √(k_eff/m) about wherever that equilibrium happens to sit. Use the supplied topic model Y_out = 100*m*g/k, where Y_out is STATIC STRETCH Δ = mg/k (cm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; g = 9.81 m s⁻². k is Spring stiffness k, measured in N m⁻¹. Keep m (Hanging mass m) = 0.3 kg; a (Amplitude A) = 5 cm fixed. For this exercise use only 24.25 N m⁻¹ ≤ k ≤ 55.75 N m⁻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 k = 31.18 N m⁻¹ and 35.59 N m⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δk across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
STATIC STRETCH Δ = mg/k against k
Predictions from the supplied Mass–Spring Systems 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): STATIC STRETCH Δ = mg/k / cm
Horizontal axis (x): k: Spring stiffness k / N m⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Spring stiffness k / N m⁻¹ | STATIC STRETCH Δ = mg/k / cm |
|---|---|
| 24.25 | 12.136082474226805 |
| 32.125 | 9.161089494163425 |
| 40 | 7.3575 |
| 47.875 | 6.147258485639687 |
| 55.75 | 5.278923766816144 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of k separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At k = 31.18 N m⁻¹, substitute into Y_out = 100*m*g/k: Y₀ = 9.43874 cm.
- At k = 35.59 N m⁻¹, the same fixed-parameter model gives Y₁ = 8.26918 cm.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.16957 cm; the input change is Δk = 4.41 N m⁻¹.
- Average sensitivity is ΔY_out/Δk = -0.265208 cm per N m⁻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 = 9.43874 cm, Y₁ = 8.26918 cm; ΔY_out = -1.16957 cm; average sensitivity = -0.265208 cm per N m⁻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.
Mass–Spring Systems: infer an input with uncertainty
A mass–spring system is a body of mass m held by one or more springs whose net force about equilibrium is −k_eff times the displacement, so it oscillates with ω = √(k_eff/m) about wherever that equilibrium happens to sit. Use the supplied topic model Y_out = 100*m*g/k, where Y_out is STATIC STRETCH Δ = mg/k (cm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; g = 9.81 m s⁻². k is Spring stiffness k, measured in N m⁻¹. Keep m (Hanging mass m) = 0.3 kg; a (Amplitude A) = 5 cm fixed. For this exercise use only 24.25 N m⁻¹ ≤ k ≤ 55.75 N m⁻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 = 6.7709651 cm, with a bounded reading uncertainty ±0.22381 cm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00315 N m⁻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.
STATIC STRETCH Δ = mg/k against k
Predictions from the supplied Mass–Spring Systems 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): STATIC STRETCH Δ = mg/k / cm
Horizontal axis (x): k: Spring stiffness k / N m⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Spring stiffness k / N m⁻¹ | STATIC STRETCH Δ = mg/k / cm |
|---|---|
| 24.25 | 12.136082474226805 |
| 32.125 | 9.161089494163425 |
| 40 | 7.3575 |
| 47.875 | 6.147258485639687 |
| 55.75 | 5.278923766816144 |
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(k) = 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(24.25) = 12.1361 cm and Y_out(55.75) = 5.27892 cm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives k_line = 24.25 + (6.77097 − 12.1361)(55.75 − 24.25)/(5.27892 − 12.1361) = 48.896 N m⁻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 k = 43.465 N m⁻¹; substituting back gives Y_out = 6.7709651 cm.
- The interpolation error in the input is |k_line − k| = 5.43095 N m⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 6.5471551 cm and 6.9947751 cm. Sort the roots even if the response decreases: 42.0743 N m⁻¹ ≤ k ≤ 44.9508 N m⁻¹.
- The lower and upper input deviations are 1.39074 N m⁻¹ and 1.48582 N m⁻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 k ≈ 43.465 N m⁻¹; compatible interval [42.0743, 44.9508] N m⁻¹. Endpoint interpolation gives 48.896 N m⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.