University Physics I · topic UPI-14.03
Displacement, velocity, and acceleration in SHM
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.
Displacement, Velocity & Acceleration in SHM: predict a finite change
In simple harmonic motion, displacement, velocity and acceleration are sinusoids sharing one angular frequency ω, with amplitudes A, ωA and ω²A and phases that step a quarter cycle ahead with each derivative. Use the supplied topic model Y_out = sqrt(x0^2+(v0/w)^2), where Y_out is AMPLITUDE A (cm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. x0 is Snapshot position x₀, measured in cm. Keep v0 (Snapshot velocity v₀) = -8 cm/s; w (Angular frequency ω) = 2 rad/s fixed. For this exercise use only -3.15 cm ≤ x0 ≤ -0.21 cm 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 x0 = -2.5032 cm and -2.0916 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δx0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
AMPLITUDE A against x0
Predictions from the supplied Displacement, Velocity & Acceleration 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): AMPLITUDE A / cm
Horizontal axis (x): x0: Snapshot position x₀ / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| x0: Snapshot position x₀ / cm | AMPLITUDE A / cm |
|---|---|
| -3.15 | 5.091414341811124 |
| -2.415 | 4.672496655964561 |
| -1.68 | 4.338478996146 |
| -0.9449999999999998 | 4.110112528873145 |
| -0.20999999999999996 | 4.005508706768717 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of x0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At x0 = -2.5032 cm, substitute into Y_out = sqrt(x0^2+(v0/w)^2): Y₀ = 4.71869 cm.
- At x0 = -2.0916 cm, the same fixed-parameter model gives Y₁ = 4.51384 cm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.204843 cm; the input change is Δx0 = 0.4116 cm.
- Average sensitivity is ΔY_out/Δx0 = -0.497675 cm per cm. 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 = 4.71869 cm, Y₁ = 4.51384 cm; ΔY_out = -0.204843 cm; average sensitivity = -0.497675 cm per cm.
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.
Displacement, Velocity & Acceleration in SHM: infer an input with uncertainty
In simple harmonic motion, displacement, velocity and acceleration are sinusoids sharing one angular frequency ω, with amplitudes A, ωA and ω²A and phases that step a quarter cycle ahead with each derivative. Use the supplied topic model Y_out = sqrt(x0^2+(v0/w)^2), where Y_out is AMPLITUDE A (cm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. x0 is Snapshot position x₀, measured in cm. Keep v0 (Snapshot velocity v₀) = -8 cm/s; w (Angular frequency ω) = 2 rad/s fixed. For this exercise use only -3.15 cm ≤ x0 ≤ -0.21 cm 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 = 4.2237855 cm, with a bounded reading uncertainty ±0.032742 cm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown x0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000294 cm, 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 x0
Predictions from the supplied Displacement, Velocity & Acceleration 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): AMPLITUDE A / cm
Horizontal axis (x): x0: Snapshot position x₀ / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| x0: Snapshot position x₀ / cm | AMPLITUDE A / cm |
|---|---|
| -3.15 | 5.091414341811124 |
| -2.415 | 4.672496655964561 |
| -1.68 | 4.338478996146 |
| -0.9449999999999998 | 4.110112528873145 |
| -0.20999999999999996 | 4.005508706768717 |
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(x0) = 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(-3.15) = 5.09141 cm and Y_out(-0.21) = 4.00551 cm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives x0_line = -3.15 + (4.22379 − 5.09141)(-0.21 − -3.15)/(4.00551 − 5.09141) = -0.800966 cm. 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 x0 = -1.3566 cm; substituting back gives Y_out = 4.2237855 cm.
- The interpolation error in the input is |x0_line − x0| = 0.555634 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.1910435 cm and 4.2565275 cm. Sort the roots even if the response decreases: -1.45534 cm ≤ x0 ≤ -1.25094 cm.
- The lower and upper input deviations are 0.0987439 cm and 0.105662 cm. 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 x0 ≈ -1.3566 cm; compatible interval [-1.45534, -1.25094] cm. Endpoint interpolation gives -0.800966 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.