AP Physics C: Mechanics:: Oscillations

What is the DIFFERENTIAL EQUATION that DEFINES simple harmonic motion (SHM), derived by substituting the spring force, F=-kx, into Newton's second law, m*(d^2x/dt^2)=F, and what mathematical PROPERTY of the SOLUTION x(t) does this equation require?

- **A)** This concept has no actual mathematical relationship between Newton's second law, the spring force, and the resulting differential equation for SHM
- **B)** The SHM differential equation involves only the FIRST derivative of x(t), rather than the second derivative
- **C)** The SHM differential equation is d^2x/dt^2 = -(k/m)*x -- this equation REQUIRES that the SECOND DERIVATIVE of x(t) be PROPORTIONAL to the NEGATIVE of x(t) itself, a defining mathematical property satisfied by SINUSOIDAL functions (sine and cosine)
- **D)** The SHM differential equation requires the second derivative to be proportional to the POSITIVE of x(t), rather than the negative, contradicting the actual restoring-force-based equation

How is it VERIFIED, using calculus, that x(t)=A*cos(omega*t) (with omega=sqrt(k/m)) is a VALID SOLUTION to the SHM differential equation, d^2x/dt^2=-(k/m)*x, by DIFFERENTIATING x(t) TWICE and substituting back into the equation?

- **A)** This concept has no actual relationship between verifying a proposed solution and differentiating it twice to check against the original differential equation
- **B)** Differentiating x(t)=A*cos(omega*t) TWICE gives d^2x/dt^2 = -A*omega^2*cos(omega*t) = -omega^2*x(t); substituting omega^2=k/m gives d^2x/dt^2=-(k/m)*x(t), EXACTLY matching the original differential equation, confirming x(t)=A*cos(omega*t) is indeed a VALID solution
- **C)** Differentiating x(t) twice produces a result that does NOT match the original differential equation, contradicting the well-established validity of this SHM solution
- **D)** This verification requires INTEGRATING (rather than differentiating) the proposed solution, contradicting the correct differentiation-based verification method

What is the GENERAL SOLUTION to the SHM differential equation, d^2x/dt^2=-(k/m)*x, expressed with BOTH an AMPLITUDE and a PHASE CONSTANT, x(t)=A*cos(omega*t+phi), and what ROLE does the PHASE CONSTANT, phi, play in determining the specific motion described?

- **A)** The phase constant ONLY affects the AMPLITUDE of the resulting motion, rather than affecting the TIMING (horizontal shift) of the oscillation
- **B)** The phase constant has no actual mathematical role in the general SHM solution, making it an unnecessary addition to the simpler x(t)=A*cos(omega*t) formula
- **C)** The GENERAL solution, x(t)=A*cos(omega*t+phi), includes a PHASE CONSTANT (phi) that shifts the cosine function horizontally in time, allowing this single general formula to describe SHM starting from ANY initial position and velocity, not just the special case starting at maximum displacement (phi=0)
- **D)** This concept has no actual relationship between the phase constant and describing SHM starting from different initial conditions

GIVEN specific INITIAL CONDITIONS, x(0)=x0 and v(0)=v0, how can the AMPLITUDE, A, and PHASE CONSTANT, phi, of the general SHM solution, x(t)=A*cos(omega*t+phi), be DETERMINED, using the position and velocity equations evaluated at t=0?

- **A)** This concept has no actual relationship between initial conditions (x0, v0) and determining the amplitude and phase constant of the general SHM solution
- **B)** Only the amplitude (not the phase constant) can be determined from initial conditions, with the phase constant remaining permanently undetermined
- **C)** Evaluating x(t) and v(t)=-A*omega*sin(omega*t+phi) at t=0 gives TWO EQUATIONS: x0=A*cos(phi) and v0=-A*omega*sin(phi); these can be SOLVED SIMULTANEOUSLY (e.g., using the Pythagorean-like relationship A^2=x0^2+(v0/omega)^2, and tan(phi)=-v0/(omega*x0)) to find both A and phi
- **D)** Amplitude and phase constant cannot actually be determined from given initial position and velocity conditions using any calculus-based method

How is the SIMPLE PENDULUM'S differential equation of motion, d^2theta/dt^2 = -(g/L)*sin(theta), DERIVED by applying NEWTON'S SECOND LAW FOR ROTATION (tau_net=I*alpha) to a pendulum, using the gravitational torque about the pivot point?

- **A)** This differential equation has no actual derivation from applying Newton's second law for rotation to a simple pendulum
- **B)** The gravitational torque about the pivot is tau=-m*g*L*sin(theta) (the restoring torque, negative since it opposes increasing theta); with rotational inertia I=m*L^2 for a simple pendulum, Newton's second law for rotation gives m*L^2*(d^2theta/dt^2) = -m*g*L*sin(theta), which simplifies to d^2theta/dt^2 = -(g/L)*sin(theta)
- **C)** The pendulum's differential equation involves POSITIVE (rather than negative) sin(theta), contradicting the actual restoring-torque-based equation
- **D)** This derivation requires using LINEAR (rather than rotational) Newton's second law, contradicting the actual rotational-dynamics-based approach needed for a pendulum

Why is the EXACT pendulum differential equation, d^2theta/dt^2=-(g/L)*sin(theta), CONSIDERABLY MORE DIFFICULT to solve than the SHM differential equation, and how does the SMALL-ANGLE APPROXIMATION, sin(theta)≈theta (valid for SMALL theta, measured in radians), SIMPLIFY this equation into a SOLVABLE SHM form?

- **A)** The small-angle approximation makes the pendulum's differential equation MORE difficult to solve, contradicting its actual simplifying effect
- **B)** This concept has no actual relationship between the small-angle approximation and simplifying the pendulum's differential equation into a solvable SHM form
- **C)** The exact pendulum equation is ALREADY LINEAR (not requiring the small-angle approximation), contradicting the actual nonlinear nature of the sin(theta) term
- **D)** The exact equation is NONLINEAR (due to the sin(theta) term), making it much harder to solve than the LINEAR SHM equation; for SMALL angles, sin(theta)≈theta (from the Taylor series approximation), which simplifies the equation to d^2theta/dt^2≈-(g/L)*theta -- EXACTLY the SHM differential equation form (with omega^2=g/L), now SOLVABLE using the familiar sinusoidal solution

How is the ROTATIONAL KINETIC ENERGY formula, KE_rot=(1/2)*I*omega^2 (from the earlier Rotation unit), APPLIED to a PHYSICAL PENDULUM (an extended, rigid object oscillating about a pivot, rather than a simple point-mass pendulum), to derive its DIFFERENTIAL EQUATION using an ENERGY-BASED (rather than torque-based) approach?

- **A)** Using energy conservation, the SUM of rotational kinetic energy, (1/2)*I*(dtheta/dt)^2, and gravitational potential energy (based on the center of mass's height) remains CONSTANT; DIFFERENTIATING this energy-conservation equation with respect to TIME (and simplifying) produces the SAME differential equation of motion that would result from the torque-based approach, providing an ALTERNATIVE derivation method
- **B)** Differentiating an energy-conservation equation with respect to time produces a result COMPLETELY UNRELATED to the object's differential equation of motion
- **C)** Energy-based methods have no actual relationship to deriving a physical pendulum's differential equation of motion
- **D)** This energy-based approach can ONLY be used for simple pendulums (point masses), with no actual applicability to extended, physical pendulums

What is a PHYSICAL PENDULUM'S PERIOD formula, T=2*pi*sqrt(I/(m*g*d)) (where I is the rotational inertia about the pivot, m is total mass, and d is the distance from the pivot to the center of mass), and how does this GENERALIZE the SIMPLE pendulum period formula, T=2*pi*sqrt(L/g)?

- **A)** The physical pendulum formula has no actual relationship to the simple pendulum formula, despite their structurally similar appearance
- **B)** The physical pendulum formula, T=2*pi*sqrt(I/(m*g*d)), REDUCES to the simple pendulum formula, T=2*pi*sqrt(L/g), when the object is treated as a POINT MASS (I=m*L^2, d=L), since substituting these values gives T=2*pi*sqrt((m*L^2)/(m*g*L)) = 2*pi*sqrt(L/g) -- exactly recovering the simple pendulum formula as a SPECIAL CASE
- **C)** The physical pendulum formula produces a COMPLETELY DIFFERENT result from the simple pendulum formula, even when applied to an idealized point mass
- **D)** This generalization only works for objects with ZERO rotational inertia, with no actual applicability to realistic, extended physical pendulums