| 5* | 1.20 | 28.4 | 11.76 | 41.4 |
(a) Identify the quantities that should be plotted on the vertical and horizontal axes to obtain a linear graph whose slope directly yields the spring constant \(k\).
(b) Using the linear regime (Trials 1–4), determine the experimental spring constant \(k_{exp}\) with appropriate SI units.
(c) Explain the physical significance of Trial 5. Why does the ratio \(F/\Delta x\) drop sharply, and what physical boundary of real springs has been breached?
AP Physics C Mechanics • Spring Networks Page 2 of 2
Plotted quantities:
Slope = \(\frac{\Delta F_g}{\Delta(\Delta x)} = k\) \([\text{N/m}]\)
Scoring Rubric: +1 pt for identifying \(F_g\) (or \(m\)) on vertical axis and \(\Delta x\) on horizontal axis. +1 pt for explicitly explaining that the slope directly equals \(k\) (or \(k/g\) if \(m\) is plotted).
(b) Calculation of Experimental Spring Constant \(k_{exp}\) [4 Points]
Using the linear data points (Trials 1 through 4, converting cm to meters):
Slope = \(\frac{F_4 - F_1}{\Delta x_4 - \Delta x_1} = \frac{7.84\,\text{N} - 1.96\,\text{N}}{0.161\,\text{m} - 0.041\,\text{m}} = \frac{5.88\,\text{N}}{0.120\,\text{m}} = 49.0\,\text{N/m}\)
Or evaluating average of local slopes: \(\bar{k} = \frac{47.8 + 49.0 + 48.2 + 48.7}{4} = 48.4\,\text{N/m}\)
Scoring Rubric: +1 pt: Converting \(\Delta x\) from \(\text{cm}\) to \(\text{m}\).
+1 pt: Using two representative points on the linear line (not single point).
+1 pt: Numerical value in acceptable range \([48.0, 49.2]\,\text{N/m}\).
+1 pt: Proper SI units (\(\text{N/m}\) or \(\text{kg/s}^2\)).
(c) Physical Significance of Trial 5 [4 Points]
In Trial 5, the extension jumps disproportionately to \(28.4\,\text{cm}\) under \(11.76\,\text{N}\), causing the ratio \(F/\Delta x\) to drop sharply from \(\approx 48.5\,\text{N/m}\) down to \(41.4\,\text{N/m}\).
Physical Explanation: The spring has exceeded its elastic limit (yield point). Beyond this boundary, irreversible plastic deformation occurs as microscopic crystalline planes within the metal slip past one another. The material softens, and Hooke's linear restoring model is no longer valid.
Scoring Rubric: +2 pts for identifying that the elastic limit (or yield strength) was breached. +2 pts for describing permanent/plastic deformation or material softening causing disproportionate elongation.
AP Physics C Mechanics • Spring Networks Solutions Page 2 of 2
All Kinetic Energy
As the mass passes through the natural resting position, the spring is neither stretched nor compressed:
Potential: \(U_s(0) = 0\)
Kinetic: \(K_{\max} = E = \frac{1}{2} m v_{\max}^2\)
Setting \(\frac{1}{2}mv_{\max}^2 = \frac{1}{2}kA^2\) yields peak speed: \(v_{\max} = \sqrt{\frac{k}{m}} A = \omega A\).
Shared Partition
At any intermediate position \(x\), total energy is distributed between both forms:
\[ K(x) = E - U_s(x) = \frac{1}{2} k (A^2 - x^2) \implies v(x) = \pm \sqrt{\frac{k}{m}\left(A^2 - x^2\right)} \]
Equal Split (\(K = U = \frac{1}{2}E\)):
Occurs at \(x = \pm \frac{A}{\sqrt{2}} \approx \pm 0.707 A\). Half energy in each form.
Half-Amplitude (\(x = \pm A/2\)):
\(U = \frac{1}{4}E\) (25%), leaving \(K = \frac{3}{4}E\) (75%). Speed is still \(86.6\%\) of maximum!
AP Physics C Mechanics • Elastic Energy Page 2 of 6
Visualizing Conservation
Part 3: Spatial Graphs
Position x Energy (J) -A +A 0 E = ½ kA² U(x) = ½ kx² K(x) = E - ½ kx² E/2 K = U = E/2
100% x = -A Turning Pt 50% 50% -0.71 A Equal Split 100% x = 0 Equilibrium 50% 50% +0.71 A Equal Split 100% x = +A Turning Pt
AP Physics C Mechanics • Elastic Energy Page 3 of 6
Temporal Dynamics
Part 4: Energy vs. Time
For an oscillator released from rest at \(x = A\) at \(t = 0\), position and velocity evolve as \(x(t) = A\cos(\omega t)\) and \(v(t) = -A\omega\sin(\omega t)\). Substituting these into the energy definitions:
Elastic Potential Energy: \(U(t) = \frac{1}{2} k x(t)^2 = \frac{1}{2} k A^2 \cos^2(\omega t)\)
Kinetic Energy: \(K(t) = \frac{1}{2} m v(t)^2 = \frac{1}{2} m (A\omega)^2 \sin^2(\omega t) = \frac{1}{2} k A^2 \sin^2(\omega t)\)
\[ E_{tot}(t) = K(t) + U(t) = \frac{1}{2} k A^2 \left[ \cos^2(\omega t) + \sin^2(\omega t) \right] \equiv \frac{1}{2} k A^2 = \text{constant} \]
Energy Frequency: \(\omega_E = 2\omega\)
Time t Energy (J) 0 T/4 T/2 3T/4 T E E/2 U(t) = ½ kA² cos²(ωt) K(t) = ½ kA² sin²(ωt) Energy Period TE = T/2 (1 Full Cycle of Energy) Motion Period T = 2π/ω (2 Full Cycles of Energy Occur!)
Why the Frequency Doubles (\(\omega_E = 2\omega\))
Using the trigonometric power-reduction identity \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\), the energy equations become:
\[ U(t) = \frac{1}{4}kA^2 [1 + \cos(2\omega t)], \quad K(t) = \frac{1}{4}kA^2 [1 - \cos(2\omega t)] \]
Energy oscillates at twice the mechanical frequency: \(f_E = 2f\).
Equipartition Time-Averages
Because the oscillating \(\cos(2\omega t)\) terms average to zero over any full cycle, the time-averaged energies are identical:
\[ \langle K \rangle = \langle U \rangle = \frac{1}{4} k A^2 = \frac{1}{2} E \]
On average over time, exactly half the total energy is kinetic and half is potential.
AP Physics C Mechanics • Elastic Energy Page 4 of 6
Vertical Systems
Part 5: Vertical Springs
When a spring hangs vertically, you have two potential energies simultaneously: elastic spring energy \(U_s\) and gravitational potential energy \(U_g\). Tracking both separately is tedious. But physics gives us an elegant shortcut: if you measure position from the hanging resting position, gravity disappears from your energy and force equations!
Vertical Coordinates Unstretched Eq: y = 0 y₀ = mg/k m +A -A 2A
Step 1: The Hanging Equilibrium Stretch
When hung, mass \(m\) stretches the spring by \(y_0\) until spring force balances gravity:
\[ k y_0 = mg \implies y_0 = \frac{mg}{k} \]
Step 2: Total Potential from New Resting Spot
Let \(y\) be displacement measured from this resting position. Total stretch is \(y_0 + y\):
\[ U_{tot} = \frac{1}{2} k (y_0 + y)^2 - mg y \] \[ U_{tot} = \frac{1}{2} k y_0^2 + (k y_0 - mg) y + \frac{1}{2} k y^2 \]
Step 3: The Cross-Term Vanishes!
Because \(ky_0 - mg = 0\), the linear gravity term cancels out completely:
\[ U_{tot}(y) = \frac{1}{2} k y^2 + \text{constant} \]
Takeaway: Treat a vertical spring exactly like a horizontal spring of stiffness \(k\) centered at the loaded resting spot!
AP Physics C Mechanics • Elastic Energy Page 5 of 6
Advanced Applications
Part 6: Advanced Wells
When springs undergo extreme deformation, restoring forces exhibit higher-order cubic terms: \(F(x) = -\alpha x - \beta x^3\).
Hardening Spring (\(\beta > 0\))
Stiffness increases with displacement. Potential well is steeper than a parabola:
\[ U(x) = \frac{1}{2}\alpha x^2 + \frac{1}{4}\beta x^4 \]
Period decreases with increasing amplitude \(A\).
Softening Spring (\(\beta < 0\))
Stiffness decreases with displacement. Potential well is shallower than a parabola:
\[ U(x) = \frac{1}{2}\alpha x^2 - \frac{1}{4}|\beta| x^4 \]
Period increases with increasing amplitude \(A\).
Find Force from Potential: Take the negative derivative: \(F_x(x) = -dU/dx\).
Find Equilibrium Points: Set \(F_x(x) = 0 \iff dU/dx = 0\). If \(d^2U/dx^2 > 0\), it is stable equilibrium.
Find Turning Points: Set total energy equal to potential energy: \(E = U(x_{tp})\). There, \(K = 0\) and the particle reverses.
Find Maximum Speed: Speed \(v_{\max}\) always occurs at the minimum of \(U(x)\): \(v_{\max} = \sqrt{\frac{2(E - U_{\min})}{m}}\).
Summary of Energy at Key Points in SHM
Ends (\(x = \pm A\)) \(K = 0, U = E\)
Equilibrium (\(x = 0\)) \(K = E, U = 0\)
In Between (\(|x| = \frac{A}{\sqrt{2}}\)) \(K = \frac{E}{2}, U = \frac{E}{2}\)
AP Physics C Mechanics • Elastic Energy Page 6 of 6
Scoring: +1 pt: Stating clear reference. +1 pt: Gravitational term with total drop \(h + y_{\max}\). +1 pt: Elastic energy term. +1 pt: Equating initial and final energy.
(b) Quadratic Solution for Maximum Compression \(y_{\max}\) [5 Points]
Rearrange into standard quadratic form \(a y^2 + b y + c = 0\):
\[ \frac{1}{2} k y_{\max}^2 - Mg y_{\max} - Mgh = 0 \iff k y_{\max}^2 - 2Mg y_{\max} - 2Mgh = 0 \] +2 pts: standard quadratic setup
Apply the quadratic formula and choose the positive root (compression > 0):
\[ y_{\max} = \frac{2Mg + \sqrt{4M^2g^2 - 4(k)(-2Mgh)}}{2k} = \frac{Mg}{k} + \sqrt{\left(\frac{Mg}{k}\right)^2 + \frac{2Mgh}{k}} \] +3 pts: algebra & physical root selection
(c) Evaluation of Maximum Speed Claim [4 Points]
The student's claim is INCORRECT.
Proof via Newton's Second Law: At the instant the block first contacts the spring (\(y = 0\)), the spring is unstretched, so \(F_s = 0\). The only force acting is gravity downward (\(\Sigma F = Mg > 0\)). Since acceleration is positive (downward), the block is still speeding up!
Maximum velocity occurs when acceleration is ZERO: \[ \Sigma F = Mg - ky = 0 \implies y_{v_{\max}} = \frac{Mg}{k} \]
Proof via Calculus (\(dK/dy = 0\)): \(K(y) = Mg(h + y) - \frac{1}{2}ky^2 \implies \frac{dK}{dy} = Mg - ky = 0 \implies y = \frac{Mg}{k}\).
Scoring: +1 pt: Refuting claim. +2 pts: Correct calculus derivation or force balance showing acceleration is zero at equilibrium. +1 pt: Stating \(y = Mg/k\).
AP Physics C Mechanics • Energy Conservation Solutions Page 2 of 2
Notice the 4 in the denominator came straight from the power rule \(\frac{x^{3+1}}{3+1} = \frac{x^4}{4}\)!
2 Find the Total Mechanical Energy \(E\) Evaluate at Amplitude
Because the block was released from rest (\(v = 0\)) at \(x = 0.20\,\text{m}\), all energy is potential:
\[ E = U(0.20) = \frac{1}{4}(200)(0.20)^4 = 50 \times (0.0016) = 0.080\,\text{J} \]
3 Find Maximum Speed at Equilibrium (\(x = 0\)) Use Energy Conservation
At \(x = 0\), potential energy is \(U(0) = 0\), so all \(0.080\,\text{J}\) becomes kinetic energy:
\[ \frac{1}{2} m v_{\max}^2 = E \implies \frac{1}{2}(0.50) v_{\max}^2 = 0.080 \implies v_{\max} = \sqrt{\frac{2(0.080)}{0.50}} = \sqrt{0.32} \approx 0.566\,\text{m/s} \]
4 Find Acceleration at the Release Point Newton’s Second Law
Just use \(a = \frac{F}{m}\):
\[ a = \frac{-b x^3}{m} = \frac{-200 (0.20)^3}{0.50} = \frac{-200(0.008)}{0.50} = \frac{-1.6}{0.50} = -3.2\,\text{m/s}^2 \]
AP Physics C Mechanics • Non-Linear Springs Page 2 of 3
Exam Strategy
Problem Protocol
When the force has two terms, simply integrate each term separately:
Step 1: Set up the integral
\(U(x) = -\int_0^x \left(-kx - cx^2\right) dx = \int_0^x \left(kx + cx^2\right) dx\)
Step 2: Apply the power rule to each term
\[ U(x) = \frac{1}{2} k x^2 + \frac{1}{3} c x^3 \]
Concrete Calculation: If \(k = 60\,\text{N/m}\) and \(c = 30\,\text{N/m}^2\), how much work is required to stretch the spring by \(0.10\,\text{m}\)?
\[ W = \Delta U = \frac{1}{2}(60)(0.10)^2 + \frac{1}{3}(30)(0.10)^3 = 30(0.01) + 10(0.001) = 0.30 + 0.01 = 0.31\,\text{J} \]
Step 1
Need Potential Energy? Integrate the force: \(U(x) = -\int F(x)\,dx\). Use the power rule!
Step 2
Need Force from Potential? Take the negative derivative: \(F(x) = -\frac{dU}{dx}\).
Step 3
Need Speed or Velocity? Use Conservation of Energy: \(E = \frac{1}{2}mv^2 + U(x)\). Solve for \(v\).
Step 4
Need Acceleration? Use Newton's Second Law: \(a = \frac{F(x)}{m}\). (Never use energy to find acceleration directly).
Key Takeaway: Non-linear springs are not harder than Hooke’s Law—they just have a different exponent when you use the power rule!
AP Physics C Mechanics • Non-Linear Springs Page 3 of 3
\[ W_{\text{ext}} = 0.180\,\text{J} + 100(0.00243)\,\text{J} = 0.180\,\text{J} + 0.243\,\text{J} = 0.423\,\text{J} \]
+2 pts: \(W = \Delta U\) formulation; +2 pts: exact evaluation \(0.423\,\text{J}\).
(c) Rebound Speed at Equilibrium [3 Points]
Total mechanical energy: \(E = 0.423\,\text{J}\). At \(x = 0\), \(U(0) = 0\):
\[ \frac{1}{2} M v^2 = E \implies v = \sqrt{\frac{2(0.423\,\text{J})}{0.50\,\text{kg}}} = \sqrt{1.692} \approx 1.30\,\text{m/s} \]
+1.5 pts: energy conservation; +1.5 pts: final velocity value with units.
(d) Inverse-Quartic Power Law Integration (\(n = -4\)) [4 Points]
\[ \Delta U = -\int_{2d}^d F(x)\,dx = -\int_{2d}^d \left(-k_0 x^{-4}\right) dx = k_0 \int_{2d}^d x^{-4} dx \]
Applying the power rule \(\int x^{-4} dx = \frac{x^{-3}}{-3} = -\frac{1}{3x^3}\):
\[ \Delta U = k_0 \left[ -\frac{1}{3x^3} \right]_{2d}^d = -\frac{k_0}{3} \left[ \frac{1}{d^3} - \frac{1}{(2d)^3} \right] = -\frac{k_0}{3} \left[ \frac{7}{8d^3} \right] = -\frac{7k_0}{24 d^3} \]
Scoring: +1 pt: integral setup. +2 pts: power rule with negative exponent \(-4\). +1 pt: evaluating fractional limits \(\frac{7k_0}{24d^3}\).
AP Physics C Mechanics • Nonlinear Springs Solutions Page 2 of 2
Isochronism: Because the differential equation is linear, period \(T\) is completely independent of amplitude \(A\).
AP Physics C Mechanics • Calculus of Oscillations Page 1 of 3
Comparative Waveforms
Phase Constant \(\phi = 0\)
Below are the synchronized profiles of position, velocity, and acceleration across one full cycle (\(0 \le t \le T\)) for an oscillator released from rest at \(x = +A\). Notice how the derivative relationships manifest visually:
x(t) +A 0 -A v(t) +ωA 0 -ωA a(t) +ω²A 0 -ω²A t = 0 T/8 T/4 3T/8 T/2 5T/8 3T/4 7T/8 T
| Time \(t\) | Position \(x(t)\) | Velocity \(v(t)\) | Acceleration \(a(t)\) | Physical Dynamic State |
|---|---|---|---|---|
| \(0\) | \(+A\) | \(0\) | \(-\omega^2 A\) | Positive Turning Point: Instantaneous Rest |
| \(T/4\) | \(0\) | \(-\omega A\) | \(0\) | Equilibrium Crossing: Max Speed in \(-x\) |
| \(T/2\) | \(-A\) | \(0\) | \(+\omega^2 A\) | Negative Turning Point: Max Restoring Force |
| \(3T/4\) | \(0\) | \(+\omega A\) | \(0\) | Equilibrium Crossing: Max Speed in \(+x\) |
| \(T\) | \(+A\) | \(0\) | \(-\omega^2 A\) | Full Cycle Completion: Restores Initial State |
Direct Proof of \(\frac{d^2x}{dt^2} = -\omega^2 x\): Compare the top curve \(x(t)\) and bottom curve \(a(t)\). At every instant in time, the acceleration wave is a perfectly inverted reflection of the position wave scaled by constant factor \(\omega^2\).
AP Physics C Mechanics • Calculus of Oscillations Page 2 of 3
Energy & Velocity Calculus
Conservation & Calculus
Substituting \(x(t) = A\cos(\omega t + \phi)\) and \(v(t) = -A\omega\sin(\omega t + \phi)\) into elastic potential and kinetic energies:
\(U(t) = \frac{1}{2} k x(t)^2 = \frac{1}{2} k A^2 \cos^2(\omega t + \phi)\) \(K(t) = \frac{1}{2} m v(t)^2 = \frac{1}{2} k A^2 \sin^2(\omega t + \phi)\)
\[ E_{\text{tot}} = K(t) + U(t) = \frac{1}{2} k A^2 \left[ \sin^2(\omega t + \phi) + \cos^2(\omega t + \phi) \right] \equiv \frac{1}{2} k A^2 = \text{constant} \]
Energy Frequency Doubling: Because \(\sin^2\theta\) and \(\cos^2\theta\) complete two cycles per oscillation cycle, energy oscillations occur at \(2\omega\) with period \(T_E = T/2\).
Energy conservation allows calculating velocity directly from position \(x\) without reference to time:
\[ \frac{1}{2} m v^2 + \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \implies v(x) = \pm \sqrt{\frac{k}{m}(A^2 - x^2)} = \pm \omega \sqrt{A^2 - x^2} \]
At Equilibrium (\(x = 0\)) Speed reaches its absolute maximum: \(v_{\max} = \omega A\). All energy is kinetic.
At Midpoint (\(x = \pm A/2\)) Speed is \(v = \frac{\sqrt{3}}{2} v_{\max} \approx 86.6\%\) of maximum. \(75\%\) of energy is kinetic.
At Turning Points (\(x = \pm A\)) Speed drops to zero: \(v = 0\). All energy is stored in elastic potential form.
Setting \(v = \frac{dx}{dt}\) in the energy relation allows direct separation of variables:
\[ \frac{dx}{dt} = \omega \sqrt{A^2 - x^2} \implies \int \frac{dx}{\sqrt{A^2 - x^2}} = \omega \int dt \implies \arcsin\left(\frac{x}{A}\right) = \omega t + \phi_0 \]
Inverting yields \(x(t) = A\sin(\omega t + \phi_0) = A\cos(\omega t + \phi)\), mathematically proving that energy conservation and Newton's Second Law define the exact same sinusoidal harmonic trajectory.
AP Physics C Mechanics • Calculus of Oscillations Page 3 of 3
| \(3T/4\) | Equilibrium Crossing (\(+x\) direction) |
(c) Using your graphs, explain both geometrically and from \(\frac{d^2x}{dt^2} = -\omega^2 x\) why the acceleration curve is an exact upside-down copy of position, and state the phase difference between \(x(t)\) and \(v(t)\). [4 pts]
AP Physics C Mechanics • Calculus of Oscillations Page 2 of 3
Energetics & Calculus
[14 Points]
Explore mechanical energy conservation, velocity as a function of position \(v(x)\), and direct integration via first-order separation of variables.
(a) Substitute \(x(t) = A\cos(\omega t + \phi)\) and \(v(t) = -A\omega\sin(\omega t + \phi)\) into \(E_{\text{tot}} = \frac{1}{2}kx^2 + \frac{1}{2}mv^2\). Use the Pythagorean trigonometric identity to prove that total energy is strictly constant in time (\(E = \frac{1}{2}kA^2\)), and explain why energy transformations oscillate at frequency \(2\omega\). [4 pts]
(b) Starting from total mechanical energy \(\frac{1}{2}kx^2 + \frac{1}{2}mv^2 = \frac{1}{2}kA^2\), algebraically isolate velocity to derive \(v(x) = \pm\omega\sqrt{A^2 - x^2}\). Then, calculate the speed of the mass when its position is at \(x = +A/2\) as an exact fraction of maximum speed \(v_{\max}\). [5 pts]
(c) Setting \(v = \frac{dx}{dt}\) in the energy relation gives \(\frac{dx}{dt} = \omega\sqrt{A^2 - x^2}\). Perform a separation of variables to integrate both sides: \(\int \frac{dx}{\sqrt{A^2 - x^2}} = \omega \int dt\). Evaluate the integral using \(\arcsin\) and solve explicitly for \(x(t)\), verifying that it reproduces the harmonic solution. [5 pts]
AP Physics C Mechanics • Calculus of Oscillations Page 3 of 3
| \(-\omega A\) |
| \(0\) |
| Equilibrium Crossing: Max Speed in \(-x\) |
| \(T/2\) | \(-A\) | \(0\) | \(+\omega^2 A\) | Negative Turning Point: Max Restoring Force |
| \(3T/4\) | \(0\) | \(+\omega A\) | \(0\) | Equilibrium Crossing: Max Speed in \(+x\) |
+1 pt for each fully correct row.
(c) Calculus Interpretation of Waveforms [4 Points]
Acceleration Inversion (+2 pts): By the standard harmonic equation \(a(t) = -\omega^2 x(t)\), acceleration is strictly proportional to negative displacement. At every instant, \(a(t)\) has the opposite algebraic sign of \(x(t)\), creating an exact vertical mirror image scaled by constant factor \(\omega^2\).
Phase Shift (+2 pts): Velocity leads position by a phase angle of \(\pi/2\) radians (a quarter-cycle \(T/4\)), achieving peak magnitudes precisely when displacement is zero.
AP Physics C Mechanics • Oscillations Exam Solutions Page 2 of 3
Energetics & Calculus [14 Points Total]
(a) Energy Conservation Proof & Frequency Doubling [4 Points]
\[ U(t) = \frac{1}{2} k x(t)^2 = \frac{1}{2} k A^2 \cos^2(\omega t + \phi) \]
\[ K(t) = \frac{1}{2} m v(t)^2 = \frac{1}{2} m (-A\omega \sin(\omega t + \phi))^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t + \phi) = \frac{1}{2} k A^2 \sin^2(\omega t + \phi) \]
\[ E_{\text{tot}} = K(t) + U(t) = \frac{1}{2} k A^2 \left[ \sin^2(\omega t + \phi) + \cos^2(\omega t + \phi) \right] \equiv \frac{1}{2} k A^2 = \text{constant} \]
Frequency Doubling (+2 pts): Identity \(\cos^2(\omega t) = \frac{1+\cos(2\omega t)}{2}\) shows energy cycles twice per mechanical oscillation (\(\omega_E = 2\omega\), \(T_E = T/2\)).
(b) Velocity as a Function of Position: \(v(x)\) [5 Points]
From total mechanical energy conservation:
\[ \frac{1}{2} m v^2 = \frac{1}{2} k A^2 - \frac{1}{2} k x^2 = \frac{1}{2} k (A^2 - x^2) \implies v(x) = \pm \sqrt{\frac{k}{m}(A^2 - x^2)} = \pm \omega \sqrt{A^2 - x^2} \]
Substitute \(x = +A/2\):
\[ |v| = \omega \sqrt{A^2 - \left(\frac{A}{2}\right)^2} = \omega \sqrt{\frac{3}{4}A^2} = \frac{\sqrt{3}}{2} \omega A = \frac{\sqrt{3}}{2} v_{\max} \approx 0.866\,v_{\max} \]
Scoring: +3 pts: Deriving \(v(x) = \pm\omega\sqrt{A^2-x^2}\). +2 pts: Evaluating \(\frac{\sqrt{3}}{2}v_{\max}\).
(c) First-Order Separation of Variables Integration [5 Points]
\[ \frac{dx}{dt} = \omega \sqrt{A^2 - x^2} \implies \int \frac{dx}{\sqrt{A^2 - x^2}} = \omega \int dt \]
Evaluate integral via inverse trigonometric definition:
\[ \arcsin\left(\frac{x}{A}\right) = \omega t + \phi_0 \implies \frac{x(t)}{A} = \sin(\omega t + \phi_0) \]
\[ x(t) = A \sin(\omega t + \phi_0) \equiv A \cos(\omega t + \phi) \]
Scoring: +2 pts: Correct variable separation. +2 pts: Evaluating \(\arcsin(x/A) = \omega t + \phi_0\). +1 pt: Inverting to verify sinusoidal harmonic trajectory.
AP Physics C Mechanics • Oscillations Exam Solutions Page 3 of 3
