Question

part a

a mass on the end of a spring undergoes simple harmonic motion. at the instant when the mass is at its equilibrium position, what is its instantaneous velocity?

at equilibrium, its instantaneous velocity is at maximum.
at equilibrium, its instantaneous velocity is zero.
instantaneous velocity cannot be determined without additional information.
at equilibrium, its instantaneous velocity is less than its maximum but not zero.

Explanation:

Analyze energy at equilibrium

Using the Conservation of Energy knowledge point

$$ E_{\text{total}} = K + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2 $$

At the equilibrium position, the displacement is zero:

$$ x = 0 \implies U = 0 $$

Determine instantaneous velocity

Using the Simple Harmonic Motion Speed knowledge point

$$ \frac{1}{2}mv^2 = \frac{1}{2}kA^2 \implies v = \pm \omega A = \pm v_{\text{max}} $$

Since all potential energy is fully converted into kinetic energy at \(x = 0\), the magnitude of the instantaneous velocity reaches its maximum value.

Answer:

• At equilibrium, its instantaneous velocity is at maximum. (Correct answer)
• At equilibrium, its instantaneous velocity is zero.
• Instantaneous velocity cannot be determined without additional information.
• At equilibrium, its instantaneous velocity is less than its maximum but not zero.