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AP Physics 1 Flashcards: Energy Of Simple Harmonic Oscillators

Study Energy Of Simple Harmonic Oscillators in AP Physics 1 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Energy Of Simple Harmonic Oscillators, giving you a quick way to review the definitions, rules, and examples that matter most for AP Physics 1.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Physics 1 Flashcards: Energy Of Simple Harmonic Oscillators

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QUESTION

Determine the amplitude if the total energy is 323232 J and k=8k = 8k=8 N/m.

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ANSWER

A=2A = 2A=2 m. Solving 32=12(8)A232 = \frac{1}{2}(8)A^232=21​(8)A2 gives A=2A = 2A=2 m.

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Flashcard 1: Determine the amplitude if the total energy is 323232 J and k=8k = 8k=8 N/m.

Answer: A=2A = 2A=2 m. Solving 32=12(8)A232 = \frac{1}{2}(8)A^232=21​(8)A2 gives A=2A = 2A=2 m.

Flashcard 2: What is the unit of potential energy in the SI system?

Answer: Joule (J). Energy is measured in joules in the SI system.

Flashcard 3: What is the formula for the total mechanical energy in a simple harmonic oscillator?

Answer: E=12kA2E = \frac{1}{2} k A^2E=21​kA2. Total energy equals half the spring constant times amplitude squared.

Flashcard 4: State the expression for potential energy in a simple harmonic oscillator.

Answer: PE=12kx2PE = \frac{1}{2} k x^2PE=21​kx2. Energy stored in spring proportional to displacement squared.

Flashcard 5: Calculate the total energy if k=150k = 150k=150 N/m and A=0.3A = 0.3A=0.3 m.

Answer: E=6.75E = 6.75E=6.75 J. Using E=12(150)(0.3)2=6.75E = \frac{1}{2}(150)(0.3)^2 = 6.75E=21​(150)(0.3)2=6.75 J.

Flashcard 6: Which type of energy is zero at the maximum displacement of a harmonic oscillator?

Answer: Kinetic energy. At maximum displacement, velocity is zero making KE zero.

Flashcard 7: What is the relationship between total energy, kinetic energy, and potential energy?

Answer: E=KE+PEE = KE + PEE=KE+PE. Conservation of energy states total equals sum of parts.

Flashcard 8: Find the spring constant if E=18E = 18E=18 J and A=0.3A = 0.3A=0.3 m.

Answer: k=400k = 400k=400 N/m. Solving 18=12k(0.3)218 = \frac{1}{2}k(0.3)^218=21​k(0.3)2 gives k=400k = 400k=400 N/m.

Flashcard 9: Identify the variable xxx in the potential energy formula of a harmonic oscillator.

Answer: Displacement from equilibrium. Distance from the equilibrium position of the oscillator.

Flashcard 10: What is the form of energy when the displacement of a harmonic oscillator is maximum?

Answer: Potential energy. At maximum displacement, velocity is zero so only PE exists.

Flashcard 11: What is the form of energy when the velocity of a harmonic oscillator is maximum?

Answer: Kinetic energy. At equilibrium, displacement is zero so only KE exists.

Flashcard 12: If E=15E = 15E=15 J, KE=9KE = 9KE=9 J at a point, find the potential energy.

Answer: PE=6PE = 6PE=6 J. Energy conservation: PE=E−KE=15−9=6PE = E - KE = 15 - 9 = 6PE=E−KE=15−9=6 J.

Flashcard 13: Determine the factor by which energy increases if amplitude is tripled.

Answer: Energy increases by a factor of 9. Energy is proportional to A2A^2A2, so tripling A increases E by 32=93^2 = 932=9.

Flashcard 14: Find the total mechanical energy if k=200k = 200k=200 N/m and A=0.5A = 0.5A=0.5 m.

Answer: E=25E = 25E=25 J. Using E=12kA2=12(200)(0.5)2=25E = \frac{1}{2}kA^2 = \frac{1}{2}(200)(0.5)^2 = 25E=21​kA2=21​(200)(0.5)2=25 J.

Flashcard 15: State the expression for kinetic energy in a simple harmonic oscillator.

Answer: KE=12mv2KE = \frac{1}{2} m v^2KE=21​mv2. Standard kinetic energy formula with mass and velocity.

Flashcard 16: What does the variable AAA represent in the energy formula of a harmonic oscillator?

Answer: Amplitude. Maximum displacement from equilibrium position.

Flashcard 17: What does the variable kkk represent in the energy formula of a harmonic oscillator?

Answer: Spring constant. The proportionality constant relating force to displacement.

Flashcard 18: What remains constant for a simple harmonic oscillator in the absence of non-conservative forces?

Answer: Total mechanical energy. Energy conservation applies when no energy is lost to friction.

Flashcard 19: Find the displacement if k=50k = 50k=50 N/m and PE=4PE = 4PE=4 J.

Answer: x=0.4x = 0.4x=0.4 m. Solving 4=12(50)x24 = \frac{1}{2}(50)x^24=21​(50)x2 gives x=0.4x = 0.4x=0.4 m.

Flashcard 20: Which type of energy is zero at the equilibrium position of a harmonic oscillator?

Answer: Potential energy. At equilibrium, displacement is zero making PE zero.

Flashcard 21: If E=10E = 10E=10 J, PE=6PE = 6PE=6 J at a point, find the kinetic energy.

Answer: KE=4KE = 4KE=4 J. Energy conservation: KE=E−PE=10−6=4KE = E - PE = 10 - 6 = 4KE=E−PE=10−6=4 J.

Flashcard 22: Calculate the mass if v=4v = 4v=4 m/s and KE=32KE = 32KE=32 J.

Answer: m=4m = 4m=4 kg. Solving 32=12m(4)232 = \frac{1}{2}m(4)^232=21​m(4)2 gives m=4m = 4m=4 kg.

Flashcard 23: Find the spring constant if the total energy is 505050 J and A=0.4A = 0.4A=0.4 m.

Answer: k=625k = 625k=625 N/m. Solving 50=12k(0.4)250 = \frac{1}{2}k(0.4)^250=21​k(0.4)2 gives k=625k = 625k=625 N/m.

Flashcard 24: Calculate the kinetic energy if m=2m = 2m=2 kg and v=3v = 3v=3 m/s.

Answer: KE=9KE = 9KE=9 J. Using KE=12mv2=12(2)(3)2=9KE = \frac{1}{2}mv^2 = \frac{1}{2}(2)(3)^2 = 9KE=21​mv2=21​(2)(3)2=9 J.

Flashcard 25: If m=0.5m = 0.5m=0.5 kg and v=5v = 5v=5 m/s, calculate the kinetic energy.

Answer: KE=6.25KE = 6.25KE=6.25 J. Using KE=12(0.5)(5)2=6.25KE = \frac{1}{2}(0.5)(5)^2 = 6.25KE=21​(0.5)(5)2=6.25 J.

Flashcard 26: Find vvv if KE=18KE = 18KE=18 J and m=3m = 3m=3 kg.

Answer: v=4v = 4v=4 m/s. Solving 18=12(3)v218 = \frac{1}{2}(3)v^218=21​(3)v2 gives v=4v = 4v=4 m/s.

Flashcard 27: What happens to the total energy if the amplitude is doubled in a harmonic oscillator?

Answer: Total energy quadruples. Energy is proportional to A2A^2A2, so doubling A increases E by 22=42^2 = 422=4.

Flashcard 28: If m=1.5m = 1.5m=1.5 kg and v=2v = 2v=2 m/s, find the kinetic energy.

Answer: KE=3KE = 3KE=3 J. Using KE=12(1.5)(2)2=3KE = \frac{1}{2}(1.5)(2)^2 = 3KE=21​(1.5)(2)2=3 J.

Flashcard 29: Determine the energy type at maximum speed in a harmonic oscillator.

Answer: Kinetic energy. Maximum speed occurs at equilibrium where all energy is kinetic.

Flashcard 30: What is the unit of amplitude in the SI system?

Answer: Meter (m). Distance is measured in meters in the SI system.