For a simple pendulum executing small-angle simple harmonic motion with time period T, which of the four plotted graphs correctly represents kinetic energy K versus time t? Take t = 0 at a turning point as shown. Image
Options

Question

For a simple pendulum executing small-angle simple harmonic motion with time period T, which of the four plotted graphs correctly represents kinetic energy K versus time t? Take t = 0 at a turning point as shown. Image

Options

G1 Physics · Accepted Answer

Step 1: A pendulum undergoing SHM has displacement x = A cos(ωt) if it starts at a turning point. Step 2: Velocity is v = −Aω sin(ωt). Step 3: Kinetic energy K = ½mv² = ½mA²ω² sin²(ωt) = K(max) sin²(ωt). Step 4: Since ω = 2π/T, K is non-negative, is zero at t = 0, T/2 and T, and has peaks at T/4 and 3T/4. Step 5: This matches the source graph D (two positive humps in one period). Final answer: Graph D (Option D).