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11 - Simple Harmonic Motion Free Online FlashCards

Study 11 - Simple Harmonic Motion with 12 free online flashcards. Review key terms, definitions, and concepts with this interactive flashcard deck.

12 cards
01
Front

What condition defines simple harmonic motion?

Back

Simple harmonic motion occurs when the restoring force is proportional to displacement and directed toward equilibrium: Fx=−kxF_x=-kx.

02
Front

What is a stable equilibrium position?

Back

The equilibrium position is where the net force is zero. A stable equilibrium also produces a restoring force when the system is displaced.

03
Front

What does amplitude mean in SHM?

Back

Amplitude AA is the maximum displacement from equilibrium, so the object moves between x=+Ax=+A and x=−Ax=-A.

04
Front

How are period and frequency related?

Back

The period is the time for one complete oscillation, while frequency is the number of oscillations per unit time: f=1Tf=\frac{1}{T}.

05
Front

What is the angular frequency of an oscillator?

Back

Angular frequency is ω=2πf=2πT\omega=2\pi f=\frac{2\pi}{T}, measured in radians per second.

06
Front

What does the phase constant determine?

Back

For x(t)=Acos⁡(ωt+ϕ)x(t)=A\cos(\omega t+\phi), the phase constant ϕ\phi determines the oscillator’s position in its cycle at a given time.

07
Front

Where is an SHM oscillator fastest?

Back

The speed is greatest at equilibrium, where x=0x=0, and the velocity is zero at the turning points, where x=±Ax=\pm A.

08
Front

Where is SHM acceleration greatest?

Back

The magnitude of acceleration is greatest at the turning points and zero at equilibrium, because a=−ω2xa=-\omega^2x.

09
Front

How does spring-oscillator energy depend on amplitude?

Back

For a spring oscillator, total mechanical energy is E=12kA2E=\frac{1}{2}kA^2. Therefore, doubling the amplitude makes the energy four times greater.

10
Front

How is energy distributed at equilibrium in a spring oscillator?

Back

At a spring oscillator’s equilibrium position, spring potential energy is zero and kinetic energy is maximum. At either turning point, kinetic energy is zero and spring potential energy is maximum.

11
Front

What determines a mass–spring period?

Back

The mass–spring period is Ts=2πmkT_s=2\pi\sqrt{\frac{m}{k}}. Increasing mm increases the period, while increasing kk decreases it.

12
Front

What is the small-angle period of a simple pendulum?

Back

For small angular displacements, a simple pendulum has period Tp=2πℓgT_p=2\pi\sqrt{\frac{\ell}{g}}.