University Physics I · unit UPI-14
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Periodic Motion & the SHM Model: infer an input with uncertainty
Simple harmonic motion is motion about a stable equilibrium in which the acceleration always points back toward equilibrium and is proportional to the displacement from it, a = −ω²x, with the same constant ω² at every displacement. Use the supplied topic model Y_out = 2*pi/sqrt(k/0.5), where Y_out is PERIOD T = 2π/ω (s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. k is Stiffness k (m = 0.50 kg), measured in N m⁻¹. Keep amp (Release amplitude A) = 6 cm; t (Time since release t) = 0.25 s fixed. For this exercise use only 5.1875 N m⁻¹ ≤ k ≤ 28.8125 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 = 1.003577 s, with a bounded reading uncertainty ±0.026381 s. 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 0.0023625 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.
PERIOD T = 2π/ω against k
Predictions from the supplied Periodic Motion & the SHM Model 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 = 2π/ω / s
Horizontal axis (x): k: Stiffness k (m = 0.50 kg) / N m⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Stiffness k (m = 0.50 kg) / N m⁻¹ | PERIOD T = 2π/ω / s |
|---|---|
| 5.1875 | 1.950679031711097 |
| 11.09375 | 1.3339074040014591 |
| 17 | 1.0775573903694988 |
| 22.906249999999996 | 0.9282989780927962 |
| 28.812499999999996 | 0.8277027519760717 |
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(5.1875) = 1.95068 s and Y_out(28.8125) = 0.827703 s. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives k_line = 5.1875 + (1.00358 − 1.95068)(28.8125 − 5.1875)/(0.827703 − 1.95068) = 25.1125 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 = 19.5987 N m⁻¹; substituting back gives Y_out = 1.003577 s.
- The interpolation error in the input is |k_line − k| = 5.51374 N m⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.977196 s and 1.029958 s. Sort the roots even if the response decreases: 18.6076 N m⁻¹ ≤ k ≤ 20.6712 N m⁻¹.
- The lower and upper input deviations are 0.991134 N m⁻¹ and 1.07248 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 ≈ 19.5987 N m⁻¹; compatible interval [18.6076, 20.6712] N m⁻¹. Endpoint interpolation gives 25.1125 N m⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The SHM Differential Equation: infer an input with uncertainty
The SHM differential equation, d²x/dt² = −ω²x, says the acceleration is always proportional to the displacement from equilibrium and directed back toward it; every solution is a sinusoid of angular frequency ω whose two constants are set by the initial position and velocity. Use the supplied topic model Y_out = hypot(x0,v0/sqrt(2*k)), 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; hypot(a,b) means √(a²+b²). k is Stiffness k (m = 0.50 kg), measured in N m⁻¹. Keep x0 (Initial position x₀) = 3 cm; v0 (Initial velocity v₀) = -32 cm s⁻¹ fixed. For this exercise use only 12.4625 N m⁻¹ ≤ k ≤ 45.5375 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 = 4.968613 cm, with a bounded reading uncertainty ±0.070401 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 0.0033075 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.
AMPLITUDE A against k
Predictions from the supplied The SHM Differential Equation 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): k: Stiffness k (m = 0.50 kg) / N m⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Stiffness k (m = 0.50 kg) / N m⁻¹ | AMPLITUDE A / cm |
|---|---|
| 12.462499999999999 | 7.076952009816638 |
| 20.731249999999996 | 5.804913037706867 |
| 28.999999999999996 | 5.162864748741062 |
| 37.26875 | 4.768443280196531 |
| 45.537499999999994 | 4.4992755692241975 |
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(12.4625) = 7.07695 cm and Y_out(45.5375) = 4.49928 cm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives k_line = 12.4625 + (4.96861 − 7.07695)(45.5375 − 12.4625)/(4.49928 − 7.07695) = 39.5153 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 = 32.6383 N m⁻¹; substituting back gives Y_out = 4.968613 cm.
- The interpolation error in the input is |k_line − k| = 6.87703 N m⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.898212 cm and 5.039014 cm. Sort the roots even if the response decreases: 31.2354 N m⁻¹ ≤ k ≤ 34.1505 N m⁻¹.
- The lower and upper input deviations are 1.40286 N m⁻¹ and 1.5122 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 ≈ 32.6383 N m⁻¹; compatible interval [31.2354, 34.1505] N m⁻¹. Endpoint interpolation gives 39.5153 N m⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
Energy in simple harmonic motion: infer an input with uncertainty
The mechanical energy of a simple harmonic oscillator is E = ½mv² + ½kx², kinetic energy plus the spring's potential energy; with no dissipation it stays fixed at ½kA² while the two parts trade places. Use the supplied topic model Y_out = sqrt(40/m)*a/100*abs(sind(th)), where Y_out is SPEED |v| (m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; abs(x) means |x|; sind/cosd/tand use degrees. m is Glider mass m, measured in kg. Keep a (Amplitude A) = 8 cm; th (Point in the cycle ωt) = 60 ° fixed. For this exercise use only 0.435937 kg ≤ m ≤ 1.81406 kg 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 = 0.38781469 m s⁻¹, with a bounded reading uncertainty ±0.0093727 m 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 m by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000137812 kg, 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.
SPEED |v| against m
Predictions from the supplied Energy in simple harmonic motion 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): SPEED |v| / m s⁻¹
Horizontal axis (x): m: Glider mass m / kg
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Glider mass m / kg | SPEED |v| / m s⁻¹ |
|---|---|
| 0.4359375 | 0.6636490846274721 |
| 0.7804687499999999 | 0.4959899253069622 |
| 1.125 | 0.41311822359545775 |
| 1.4695312499999997 | 0.36146079662393926 |
| 1.8140625 | 0.3253302833762233 |
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(m) = 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(0.435937) = 0.663649 m s⁻¹ and Y_out(1.81406) = 0.32533 m s⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m_line = 0.435937 + (0.387815 − 0.663649)(1.81406 − 0.435937)/(0.32533 − 0.663649) = 1.55954 kg. 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 m = 1.27659 kg; substituting back gives Y_out = 0.38781469 m s⁻¹.
- The interpolation error in the input is |m_line − m| = 0.282942 kg. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.37844199 m s⁻¹ and 0.39718739 m s⁻¹. Sort the roots even if the response decreases: 1.21706 kg ≤ m ≤ 1.34061 kg.
- The lower and upper input deviations are 0.0595384 kg and 0.0640167 kg. 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 m ≈ 1.27659 kg; compatible interval [1.21706, 1.34061] kg. Endpoint interpolation gives 1.55954 kg.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.