This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a simple pendulum at small angles, the period T is determined by T = 2π√(L/g), where L is the pendulum length and g is gravitational acceleration. Choice B is correct because it shows the standard pendulum formula T = 2π√(L/g), which with L = 2.0 m and g = 9.8 m/s² gives T = 2π√(2.0/9.8) = 2.84 s. Choice A is incorrect because it's missing the factor of 2 in front of π, which would give half the actual period. To help students: Derive the pendulum formula from the restoring torque τ = -mgL sin θ ≈ -mgLθ for small angles. Practice calculating periods for different pendulum lengths to build intuition about the √L dependence.

This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, the frequency f is determined by f = (1/2π)√(k/m), where k is the spring constant and m is mass. Choice A is correct because it properly shows f = (1/2π)√(k/m), which with k = 50 N/m and m = 0.80 kg gives f = (1/2π)√(50/0.80) = 1.26 Hz. Choice C is incorrect because it inverts the k/m ratio, resulting in incorrect units and a physically meaningless result. To help students: Connect frequency to the period formula by showing f = 1/T = 1/(2π√(m/k)) = (1/2π)√(k/m). Use unit analysis to verify that √(k/m) has units of s⁻¹, making the frequency have proper Hz units.

This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a lightly damped oscillator, the period remains approximately T ≈ 2π√(m/k), with damping primarily affecting amplitude decay rather than period. Choice A is correct because light damping causes exponential amplitude decay while leaving the period nearly unchanged - the actual period is slightly longer but the difference is negligible for light damping. Choice B is incorrect because damping doesn't make period undefined; the system still oscillates with a well-defined period. To help students: Introduce the damped oscillator equation and show that for light damping (ζ << 1), the period T ≈ T₀ = 2π√(m/k). Use energy arguments to explain why damping reduces amplitude but minimally affects frequency.

This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, the period T is determined by T = 2π√(m/k), where m is mass and k is the spring constant. Choice B is correct because it shows the proper formula T = 2π√(m/k), which when calculated with m = 0.50 kg and k = 200 N/m gives T = 2π√(0.50/200) = 0.314 s. Choice A is incorrect because it inverts the mass and spring constant ratio, which would give units of s⁻¹ rather than seconds. To help students: Emphasize dimensional analysis to verify that √(m/k) has units of time. Practice deriving the period formula from F = -kx and Newton's second law to build deeper understanding.

This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, the period T is determined by T = 2π√(m/k), where m is mass and k is the spring constant. Choice C is correct because when mass doubles (m' = 2m), the new period becomes T' = 2π√(2m/k) = √2 × 2π√(m/k) = √2 × T, showing that period increases by a factor of √2. Choice A is incorrect because it assumes a linear relationship between mass and period, when the actual relationship involves a square root. To help students: Emphasize that period depends on √m, not m directly. Practice problems where system parameters change to reinforce how T scales with √(m/k).

This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a simple pendulum, the period T is given by T = 2π√(L/g), where L is length and g is gravitational acceleration. Choice A is correct because it properly converts period to frequency using f = 1/T = 1/(2π√(L/g)) = (1/2π)√(g/L), showing understanding of the inverse relationship. Choice B is incorrect because it represents the period formula, not frequency, demonstrating confusion between these reciprocal quantities. To help students: Emphasize the distinction between period (time per cycle) and frequency (cycles per time). Practice converting between T and f using f = 1/T consistently across different SHM systems.

This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, the period T is determined by T = 2π√(m/k), showing that period is proportional to √m. Choice B is correct because when mass doubles from m to 2m, the new period T' = 2π√(2m/k) = √2 × 2π√(m/k) = √2 × T. Choice A is incorrect because it assumes a linear relationship between mass and period, when the actual relationship involves the square root. To help students: Emphasize that T ∝ √m, not T ∝ m, which means doubling mass increases period by √2 ≈ 1.414. Practice problems where students predict how period changes when mass is multiplied by various factors (4, 9, 1/4) to reinforce the square root relationship.

This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a vertical mass-spring system, the period T is still determined by T = 2π√(m/k), independent of gravity's effect on equilibrium position. Choice A is correct because the period formula T = 2π√(m/k) applies to both horizontal and vertical springs, giving T = 2π√(0.60/240) = 0.314 s. Choice C is incorrect because it incorrectly includes gravity (mg/k), which only affects the equilibrium position, not the period of oscillation. To help students: Emphasize that gravity shifts the equilibrium position but doesn't change the period. Show that the restoring force F = -k(x - x₀) still leads to the same differential equation and period formula.

This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a simple pendulum, the period is T = 2π√(L/g), so the frequency f = 1/T = 1/(2π√(L/g)) = (1/2π)√(g/L). Choice A is correct because it properly shows f = (1/2π)√(g/L), which with L = 1.6 m and g = 9.8 m/s² gives f = (1/2π)√(9.8/1.6) = 0.394 Hz. Choice B is incorrect because it inverts the g/L ratio, which would give incorrect units and violate the physics - longer pendulums have lower frequencies. To help students: Show the algebraic manipulation from T to f explicitly. Use dimensional analysis to verify that √(g/L) has units of s⁻¹, confirming the frequency formula is correct.

This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, the frequency f is given by f = (1/2π)√(k/m), where k is the spring constant and m is mass. Choice B is correct because when the spring constant is halved (k' = k/2), the new frequency becomes f' = (1/2π)√(k/2m) = (1/√2) × (1/2π)√(k/m) = f/√2, showing frequency decreases by a factor of √2. Choice C is incorrect because it assumes a linear relationship between spring constant and frequency, missing the square root dependence. To help students: Emphasize that frequency depends on √k, not k directly. Use dimensional analysis to verify that √(k/m) has units of 1/time, appropriate for frequency.