Question

A 0.350-kg mass oscillates on the end of a thin vertical spring (spring constant 265 N/m) with an amplitude of 28.0 cm. The mass passes through the equilibrium point (y=0)(y=0) with positive velocity at t=0t=0.

(a) What equation describes this motion as a function of time?

Answer 1

The equation that describes the motion of a 0.350-kg mass oscillating on a spring with a spring constant of 265 N/m and an amplitude of 28.0 cm is y(t) = 0.28 cos((√(265/0.350))t), assuming it passes through the equilibrium with positive velocity at t = 0.

Step-by-step explanation:

The motion of a 0.350-kg mass attached to a spring can be described by simple harmonic motion. Given the parameters of the oscillating system, an equation that describes its motion as a function of time can be derived using the formula:

y(t) = A cos(ωt + φ),

where:

A is the amplitude of the motion,

ω is the angular frequency, calculated as ω = √(k/m),

t is time, and

φ is the phase shift.

For the given system, as the mass passes through the equilibrium point with positive velocity at t = 0, we have:

y(t) = 0.28 cos((√(265/0.350))t)

This equation indicates that the mass starts its motion from the maximum displacement (amplitude) and moves towards the equilibrium point as time progresses.