AP Physics 1 Unit 6: Simple Harmonic Motion — Free Review Games.

SHM occurs when an object oscillates about an equilibrium position with a restoring force proportional to displacement (F = -kx).

Hooke's Law gives the restoring force of a spring: F = -kx, where k is the spring constant and x is displacement.

Velocity is maximum at the equilibrium position where all energy is kinetic and displacement is zero.

Amplitude is the maximum displacement from the equilibrium position during oscillation.

For small angles, the period of a simple pendulum is \(T = 2\pi\sqrt{L/g}\), depending only on length and gravity.

The period of a mass-spring oscillator is \(T = 2\pi\sqrt{m/k}\), depending on mass and spring constant.

\(T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.5/200} = 2\pi\sqrt{0.0025} = 2\pi(0.05) = 0.31\) s.

At maximum displacement, velocity is zero so KE = 0, and all energy is stored as potential energy.

Maximum velocity v_max = A*omega. If amplitude doubles, v_max doubles (omega depends only on m and k, not amplitude).

\(T = 2\pi\sqrt{L/g}\). With \(g\) reduced by 6, \(T\) increases by \(\sqrt{6}\) since \(T\) is proportional to \(1/\sqrt{g}\).

\(\frac{1}{2}kx^2 = \frac{1}{2}mv^2\). \(v = x\sqrt{k/m} = 0.2\sqrt{100/0.5} = 0.2\sqrt{200} = 0.2(14.14) = 2.83\) m/s.

Velocity is the derivative of position: v = dx/dt = -A*omega*sin(omega*t).

In parallel, $k_{eff} = k_1 + k_2 = 300$ N/m. \(T = 2\pi\sqrt{m/k} = 2\pi\sqrt{3/300} = 2\pi\sqrt{0.01} = 2\pi(0.1) = 0.63\) s.

When \(KE = PE\), each equals half the total energy. \(\frac{1}{2}kx^2 = \frac{1}{2}(\frac{1}{2}kA^2)\), so \(x^2 = A^2/2\), \(x = A/\sqrt{2}\).

Each cycle retains 90% of energy. After 5 cycles: (0.9)^5 = 0.59 or 59% of original energy remains.

The restoring force in SHM is proportional to displacement from equilibrium, \(F=-kx\), so it vanishes exactly where \(x=0\), the equilibrium position. The distractor 'At maximum displacement' is wrong because that is precisely where the restoring force is strongest, not zero. Students should remember that force and displacement are directly linked through Hooke's law in SHM.

By definition, a restoring force acts opposite to the displacement, always pulling or pushing the object back toward equilibrium, which is what produces oscillation. The choice 'points away from equilibrium' describes a destabilizing force, not a restoring one, and would cause the system to accelerate away rather than oscillate. This sign relationship, \(F=-kx\), is the defining feature of any SHM system.

Frequency is defined as the number of complete cycles per second, measured in Hertz, and is the reciprocal of the period. Period, in contrast, measures the time for one complete cycle, so it answers a different question about duration rather than rate. Recognizing that \(f=1/T\) is essential for converting between these two related but distinct quantities.

The period of a simple pendulum is given by \(T=2\pi\sqrt{L/g}\), which depends only on the length \(L\) and gravitational acceleration \(g\), not on mass. The distractor 'Mass of the bob and length' incorrectly includes mass, which cancels out of the equation of motion for a pendulum. This mass-independence is a key AP concept often tested by comparing pendulums of different bob masses but equal length.

As the mass moves toward equilibrium, the spring's displacement \(x\) decreases, so the elastic potential energy \(\frac{1}{2}kx^2\) decreases while kinetic energy correspondingly increases. The option stating potential energy 'remains constant' contradicts the energy conservation principle governing SHM, where energy continuously converts between forms. Students should picture SHM as a continuous exchange between kinetic and potential energy with total mechanical energy conserved.

At maximum displacement, all the mechanical energy is stored as potential energy, so kinetic energy and therefore speed must be zero at that instant. The equilibrium position is actually where speed is maximum since all energy there is kinetic, making that distractor incorrect. This turning-point behavior mirrors that of any oscillator, from springs to pendulums, at the extremes of motion.

Differentiating \(x=A\cos(\omega t)\) gives \(v=-A\omega\sin(\omega t)\), which is a sine function shifted \(90^\circ\) ahead of the cosine displacement, so velocity leads displacement. The option stating they are 'in phase' is wrong because displacement is zero exactly when velocity is at a maximum, not simultaneously extreme. This quarter-cycle phase shift between position and velocity is a hallmark of all SHM systems.

Angular frequency \(\omega\) describes how quickly the phase angle \(\omega t\) increases, measured in radians per second, and relates to frequency by \(\omega=2\pi f\). The distractor describing 'number of oscillations per second' actually defines ordinary frequency \(f\), not angular frequency, a common point of confusion. Students must keep \(\omega\), \(f\), and \(T\) distinct while remembering the conversion factor of \(2\pi\) between them.

SHM approximation for a pendulum relies on the small-angle approximation \(\sin\theta \approx \theta\), which holds accurately only for angles roughly under \(15^\circ\), making the restoring torque linear in displacement. Requiring the bob to have 'negligible mass' is irrelevant since the period of an ideal pendulum is independent of mass regardless of angle. This small-angle condition is a critical limitation students must recall when analyzing large-amplitude pendulum problems on the AP exam.

Angular frequency for a mass-spring system is \(\omega=\sqrt{k/m}=\sqrt{50/2}=\sqrt{25}=5\,\text{rad/s}\). The option \(25\,\text{rad/s}\) mistakenly omits the square root and simply reports \(k/m\), a common calculation error. Always remember to take the square root when computing \(\omega\) from \(k\) and \(m\).

Using \(T=2\pi\sqrt{L/g}\), solving for \(L\) gives \(L=g(T/2\pi)^2=9.8\times(2/6.283)^2\approx0.99\,\text{m}\). The distractor \(1.98\,\text{m}\) results from forgetting to square the ratio \(T/2\pi\) before multiplying by \(g\). Always isolate and square the correct term when rearranging the pendulum period formula.

Since \(T=2\pi\sqrt{m/k}\), quadrupling \(m\) increases \(T\) by a factor of \(\sqrt{4}=2\), so the period doubles. The distractor 'quadruples' incorrectly assumes a linear relationship between \(T\) and \(m\) rather than a square-root relationship. Remember that period depends on the square root of mass, not mass directly, for spring systems.

At equilibrium, the spring force balances gravity, so \(k=mg/x=(2)(9.8)/0.1=196\,\text{N/m}\). The option \(20\,\text{N/m}\) results from using \(g\approx10\) but then also mistakenly dividing incorrectly, giving an inconsistent value. Setting \(kx=mg\) at static equilibrium is the standard method for finding an unknown spring constant from a hanging mass.

Maximum speed in SHM is \(v_{max}=A\omega\), so \(\omega=v_{max}/A=2/0.5=4\,\text{rad/s}\). The distractor \(1\,\text{rad/s}\) incorrectly multiplies \(A\) and \(v_{max}\) rather than dividing. This relationship, \(v_{max}=A\omega\), is essential for connecting kinematic extremes to the SHM parameters.

Since \(T\propto\sqrt{L}\), the period ratio equals the square root of the length ratio, so \(\sqrt{4/1}=2\), giving a period ratio of \(2:1\). The distractor \(4:1\) incorrectly assumes period scales linearly with length rather than with its square root. This square-root dependence is a frequently tested relationship for comparing pendulums of different lengths.

By energy conservation, \(\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2\), so \(v_{max}=A\sqrt{k/m}=0.1\sqrt{400/1}=0.1\times20=2\,\text{m/s}\). The distractor \(4\,\text{m/s}\) comes from forgetting to multiply by amplitude after taking the square root of \(k/m\). This energy method, equating spring potential energy at maximum displacement to kinetic energy at equilibrium, is a core problem-solving tool in SHM.

Since \(T=2\pi\sqrt{L/g}\), reducing \(g\) to one-fourth increases \(T\) by a factor of \(\sqrt{4}=2\), meaning each swing takes twice as long and the clock runs slow. The distractor claiming the period 'remains the same' ignores the direct dependence of pendulum period on gravitational acceleration. This inverse square-root relationship between \(T\) and \(g\) is commonly tested in extraterrestrial pendulum scenarios.

Setting \(\frac{1}{2}kx^2=\frac{1}{2}(\frac{1}{2}kA^2)\) gives \(x^2=A^2/2\), so \(x=A/\sqrt{2}\), the displacement where energy is split equally between kinetic and potential forms. The distractor \(x=A/2\) is a common but incorrect guess that does not satisfy the actual energy balance equation. This equal-energy condition is a useful checkpoint problem type that tests understanding of the quadratic dependence of energy on displacement.

Frequency for a spring system is \(f=\frac{1}{2\pi}\sqrt{k/m}\), so doubling \(k\) increases \(f\) by a factor of \(\sqrt{2}\) since frequency depends on the square root of \(k\). The distractor stating frequency 'increases by a factor of \(2\)' incorrectly assumes a linear relationship between \(f\) and \(k\). Remembering that frequency and period both involve square-root dependence on stiffness and mass is vital for scaling problems.

At equilibrium, displacement \(x=0\), so by Hooke's law \(F=-kx=0\), meaning the net force and therefore acceleration are zero at that instant. The distractor claiming acceleration is 'maximum because velocity is maximum' confuses velocity and acceleration, which are out of phase in SHM. Students should remember that acceleration in SHM is proportional to displacement, not velocity, via \(a=-\omega^2x\).

As the pendulum swings from maximum angle toward the bottom, the angular displacement decreases, and since restoring torque is proportional to displacement, the torque decreases toward zero at the lowest point. The distractor stating speed 'decreases continuously' is incorrect because speed actually increases as the bob swings from the release point toward the bottom due to conversion of potential to kinetic energy. This mirrors the general SHM rule that restoring force (or torque) diminishes as the system approaches equilibrium.

Since acceleration in SHM is \(a=-\omega^2x\), it is directly proportional to displacement, so acceleration equal to half its maximum value occurs exactly at \(x=A/2\). The distractor \(x=A/\sqrt{2}\) confuses this linear proportionality with the quadratic relationship used for energy problems. Students should remember that unlike energy, acceleration and displacement in SHM are linearly related.

Gravity in a vertical spring system simply shifts the equilibrium point to where \(kx_0=mg\), but the net restoring force about this new equilibrium is still \(-kx'\), so the period formula \(T=2\pi\sqrt{m/k}\) remains identical to the horizontal case. The distractor claiming the period 'increases' incorrectly assumes gravity contributes an additional restoring term beyond the spring force. This principle, that gravity shifts equilibrium but does not alter oscillation period for linear springs, is a frequently tested AP concept.

The pendulum's period depends on \(g\) through \(T=2\pi\sqrt{L/g}\), so a smaller \(g\) on the Moon increases its period, while the spring-mass period \(T=2\pi\sqrt{m/k}\) has no \(g\) dependence and stays the same. The distractor claiming both periods 'stay the same' is wrong for the pendulum since gravity is fundamental to its restoring force mechanism. This contrast highlights a key distinction in AP Physics 1: gravity affects pendulums but never affects ideal spring-mass oscillators.

Total mechanical energy in SHM is \(E=\frac{1}{2}kA^2\), so it scales with the square of the amplitude; tripling \(A\) multiplies energy by \(3^2=9\), giving \(9E\). The distractor \(3E\) incorrectly assumes a linear relationship between energy and amplitude rather than the correct quadratic one. This quadratic dependence of energy on amplitude is a critical relationship for solving energy-scaling problems in SHM.

Since \(T=2\pi\sqrt{m/k}\) and the mass doubles while \(k\) stays constant, the new period becomes \(T'=2\pi\sqrt{2m/k}=T\sqrt{2}\). The distractor \(2T\) incorrectly assumes period scales linearly with mass instead of with its square root. This scenario tests the square-root mass dependence of spring period along with recognizing that momentum, not energy, is conserved in the sudden mass addition.

For amplitudes beyond the small-angle regime, the restoring torque \(\sin\theta\) is less than \(\theta\), weakening the true restoring effect compared to the linear approximation, which causes the actual period to be slightly longer than \(2\pi\sqrt{L/g}\). The distractor claiming the period is 'exactly equal' ignores that the small-angle approximation is only an approximation, breaking down noticeably by \(30^\circ\). Students should recognize that real pendulum period slightly increases with amplitude beyond the idealized SHM approximation.

Setting \(x(0)=A\cos\phi=A/2\) gives \(\cos\phi=1/2\), so \(\phi=\pm\pi/3\); checking the velocity \(v(0)=-A\omega\sin\phi\) must be negative, which requires \(\sin\phi>0\), satisfied by \(\phi=\pi/3\). The distractor \(-\pi/3\) gives \(\sin\phi<0\), producing a positive velocity, which contradicts the given negative-direction motion. This problem illustrates how both position and velocity conditions are needed together to uniquely determine the phase constant.

Springs in parallel combine as $k_{eff}=k+k=2k$, so the new period is \(T'=2\pi\sqrt{m/2k}=\frac{T}{\sqrt{2}}\), meaning the period is reduced by a factor of \(\sqrt{2}\). The distractor 'increased by a factor of \(\sqrt{2}\)' incorrectly assumes stiffer combined springs would slow the oscillation, when in fact greater stiffness speeds it up. Remembering that parallel springs add their constants while series springs add their compliances (reciprocals) is essential for combined-spring problems.

With \(KE=3PE\) and \(KE+PE=E=\frac{1}{2}kA^2\), substituting gives \(4PE=E\), so \(PE=E/4\), meaning \(\frac{1}{2}kx^2=\frac{1}{2}kA^2/4\), giving \(x^2=A^2/4\) and \(x=A/2\). The distractor \(x=A\sqrt{3}/2\) actually corresponds to the case where potential energy is three times kinetic energy, a reversed condition from what was asked. This problem type tests careful algebraic manipulation of the SHM energy conservation equation under specific energy ratios.

Since the pendulum's period depends on \(g\) through \(T=2\pi\sqrt{L/g}\), a decrease in \(g\) increases \(T\), causing the pendulum to swing more slowly and the clock to run slow, while the spring-mass period \(T=2\pi\sqrt{m/k}\) is unaffected by gravity and remains accurate. The distractor stating 'both clocks will run slow by the same amount' incorrectly assumes both oscillator types share the same gravitational dependence. This distinction is a classic conceptual test of whether students understand which SHM systems depend on \(g\).

Average speed equals total distance divided by total time; over one quarter-period the mass travels a distance \(A\) (from maximum displacement to equilibrium) in time \(T/4\), giving average speed \(A/(T/4)=4A/T\). The distractor involving \(\pi A/T\) confuses average speed with a more complex sinusoidal averaging technique that is unnecessary here since distance and time are both straightforward over this specific quarter-cycle interval. This distance-over-time approach works cleanly because a quarter period always spans exactly one amplitude of travel in SHM.

Adding a constant force shifts the equilibrium point to a new location where \(kx_0=F\), but the net force about this new equilibrium is still linear in the displacement from it, \(-k(x-x_0)\), so the oscillation remains SHM with the same period \(T=2\pi\sqrt{m/k}\). The distractor claiming the motion 'is no longer periodic' incorrectly assumes any additional constant force destroys the harmonic nature, when in fact only the reference point of oscillation changes. This principle, that constant external forces shift equilibrium without changing period, parallels the gravity effect on vertical springs.

By direct analogy to the mass-spring system where \(T=2\pi\sqrt{m/k}\), rotational inertia \(I\) plays the role of mass and torsion constant \(\kappa\) plays the role of spring constant, giving \(T=2\pi\sqrt{I/\kappa}\). The distractor \(T=2\pi\sqrt{\kappa/I}\) inverts the ratio, which would incorrectly predict the period decreases as rotational inertia increases. Recognizing this rotational-translational analogy helps extend SHM concepts beyond simple springs and pendulums to torsional systems.

At the lowest point of a vertical SHM, the mass is furthest below equilibrium, so the spring stretch is greatest, producing a large upward spring force that exceeds gravity, giving a net restoring force directed upward toward equilibrium. The distractor claiming 'net force is zero' incorrectly describes the equilibrium position, not the extreme lowest point, where the restoring force is actually at its maximum. This exemplifies how the net force in SHM always points toward equilibrium and is strongest at the amplitude extremes, whether the spring is horizontal or vertical.

Amplitude is specifically defined as the maximum distance the object travels from equilibrium during SHM. The distractor 'wavelength' refers to a spatial property of waves, not a single oscillator's displacement, and does not apply directly to a mass-spring or pendulum system. Recognizing amplitude as a displacement quantity, not a time or spatial-wave quantity, is fundamental for SHM vocabulary.

The spring itself provides the restoring force according to Hooke's law, \(F=-kx\), which always pulls or pushes the mass back toward equilibrium. The distractor 'air resistance' actually opposes motion generally and dissipates energy rather than providing the primary restoring mechanism that drives oscillation. Students should always identify Hooke's law as the source of the restoring force in ideal spring systems.

The restoring force for a pendulum comes from the tangential component of gravity, \(mg\sin\theta\), which acts along the arc of motion and pulls the bob back toward the lowest point. The distractor 'tension in the string' actually acts along the radial direction and provides centripetal force, not the tangential restoring force responsible for oscillation. Distinguishing the tangential (restoring) and radial (centripetal) components of forces is key to correctly analyzing pendulum motion.

SHM displacement follows \(x(t)=A\cos(\omega t+\phi)\), which is inherently sinusoidal, oscillating smoothly between positive and negative amplitude values over time. The distractor 'a parabola opening upward' describes constant-acceleration motion under a fixed force, not the periodically reversing acceleration characteristic of SHM. Recognizing the sinusoidal shape of position-time graphs is a foundational skill for interpreting SHM graphs on the AP exam.

In an ideal, frictionless SHM system, energy continuously converts between kinetic and potential forms, but the total mechanical energy \(E=\frac{1}{2}kA^2\) remains constant throughout the motion. The distractor 'it steadily decreases' describes a damped system experiencing energy loss due to friction or resistance, which is not the case for an ideal oscillator. This conservation of total mechanical energy is the basis for many SHM energy calculations.

At the highest points of a pendulum's swing, the bob momentarily stops moving, so its kinetic energy is zero while gravitational potential energy, dependent on height, is at its maximum. The distractor 'kinetic energy is maximum and potential energy is zero' actually describes the condition at the lowest point of the swing, not the highest points. This kinetic-potential energy exchange is analogous to the spring system and reflects conservation of mechanical energy throughout the swing.

The spring constant relates force to displacement via \(F=kx\), so its units must be force divided by distance, giving newtons per meter (N/m). The distractor 'N\(\cdot\)m' represents units of torque or energy, not the ratio defining spring stiffness. Keeping consistent units when computing \(k\), \(F\), and \(x\) is essential to avoid calculation errors in SHM problems.

An object passes through equilibrium twice per full cycle (once moving each direction), so successive equilibrium crossings occur every half period, \(T/2=4/2=2\,\text{s}\). The distractor \(4\,\text{s}\) mistakenly treats the full period as the time between equilibrium crossings rather than recognizing there are two crossings per cycle. Visualizing the full sinusoidal cycle helps clarify how often specific positions are revisited during SHM.

Since \(T=2\pi\sqrt{m/k}\), reducing the mass to one-fourth decreases the period by a factor of \(\sqrt{4}=2\), giving a new period of \(T/2\). The distractor \(T/4\) incorrectly assumes a direct linear relationship between mass and period instead of the actual square-root dependence. This scaling relationship is one of the most frequently tested numerical concepts in AP Physics 1 SHM problems.

Setting \(\frac{1}{2}kx^2=\frac{1}{4}(\frac{1}{2}kA^2)\) leads to \(x^2=A^2/4\), so \(x=A/2\), the displacement at which potential energy is exactly one-fourth of the total mechanical energy. The distractor \(x=A/\sqrt{2}\) actually corresponds to the case where potential energy equals half the total energy, a different scenario from the one asked. Careful algebra with the quadratic energy-displacement relationship is essential to avoid confusing similar-looking energy fraction problems.