University Physics II · topic UPII-11.05
Sinusoidal AC and rms values
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.
Sinusoidal AC and rms values: predict a finite change
The rms value of an alternating voltage or current is the square root of its mean square over a whole cycle: the steady DC value that would dissipate the same average power in the same resistor. Use the supplied topic model Y_out = 1000/f, where Y_out is PERIOD T (ms). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f is Frequency f, measured in Hz. Keep v0 (Peak voltage V₀) = 325 V; phi (Phase constant φ) = 45 ° fixed. For this exercise use only 38.2813 Hz ≤ f ≤ 136.719 Hz 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 f = 59.9375 Hz and 73.7188 Hz. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δf across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PERIOD T against f
Predictions from the supplied Sinusoidal AC and rms values 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 T / ms
Horizontal axis (x): f: Frequency f / Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Frequency f / Hz | PERIOD T / ms |
|---|---|
| 38.28125 | 26.122448979591837 |
| 62.890625 | 15.900621118012422 |
| 87.5 | 11.428571428571429 |
| 112.109375 | 8.9198606271777 |
| 136.71875 | 7.314285714285714 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of f separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At f = 59.9375 Hz, substitute into Y_out = 1000/f: Y₀ = 16.684 ms.
- At f = 73.7188 Hz, the same fixed-parameter model gives Y₁ = 13.5651 ms.
- Signed change: ΔY_out = Y₁ − Y₀ = -3.11898 ms; the input change is Δf = 13.7813 Hz.
- Average sensitivity is ΔY_out/Δf = -0.22632 ms per Hz. 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 = 16.684 ms, Y₁ = 13.5651 ms; ΔY_out = -3.11898 ms; average sensitivity = -0.22632 ms per Hz.
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.
Sinusoidal AC and rms values: infer an input with uncertainty
The rms value of an alternating voltage or current is the square root of its mean square over a whole cycle: the steady DC value that would dissipate the same average power in the same resistor. Use the supplied topic model Y_out = 1000/f, where Y_out is PERIOD T (ms). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f is Frequency f, measured in Hz. Keep v0 (Peak voltage V₀) = 325 V; phi (Phase constant φ) = 45 ° fixed. For this exercise use only 38.2813 Hz ≤ f ≤ 136.719 Hz 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 = 10.17003 ms, with a bounded reading uncertainty ±0.42836 ms. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00984375 Hz, 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 T against f
Predictions from the supplied Sinusoidal AC and rms values 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 T / ms
Horizontal axis (x): f: Frequency f / Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Frequency f / Hz | PERIOD T / ms |
|---|---|
| 38.28125 | 26.122448979591837 |
| 62.890625 | 15.900621118012422 |
| 87.5 | 11.428571428571429 |
| 112.109375 | 8.9198606271777 |
| 136.71875 | 7.314285714285714 |
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(f) = 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(38.2813) = 26.1224 ms and Y_out(136.719) = 7.31429 ms. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f_line = 38.2813 + (10.17 − 26.1224)(136.719 − 38.2813)/(7.31429 − 26.1224) = 121.772 Hz. 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 f = 98.3281 Hz; substituting back gives Y_out = 10.17003 ms.
- The interpolation error in the input is |f_line − f| = 23.4443 Hz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 9.74167 ms and 10.59839 ms. Sort the roots even if the response decreases: 94.354 Hz ≤ f ≤ 102.652 Hz.
- The lower and upper input deviations are 3.97417 Hz and 4.32368 Hz. 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 f ≈ 98.3281 Hz; compatible interval [94.354, 102.652] Hz. Endpoint interpolation gives 121.772 Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.