Question

Problem 4: A mass m = 1.5 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 95 N/m and negligible mass. The mass undergoes simple harmonic motion when placed in vertical motion, with its position given as a function of time by y(t) = A cos(ωt – φ), with the positive y-axis pointing upward. At time t = 0 the mass is observed to be at a distance d = 0.35 m below its equilibrium height with an upward speed of v0 = 6 m/s.

Answer 1

Answer:

y (t) = 0.754 * cos ( 7.96 t - 69.52)

Explanation:

Given:

m = 1.5 kg , k = 95 N / m , v₀ = 6 m / s , d = 0.35 m , t = 0

y (t) = A * cos ( ω * t - φ )

Using the equation that describe the motion

m * v = - k * x  ⇒ m * x'' = - k * x

Angular velocity is equal to

ω = √ k / m   ⇒ ω = √ 95 N /m  /  1.5 kg  

ω = 7.96  rad /s

A = v / ω   ⇒  A = 6 m /s   /   7.96 rad / s

A = 0.754

d = cos * φ    ⇒  φ = cos ⁻¹ * 0.35

φ = 69.52

y (t) = A * cos ( ω * t - φ )    ⇒  y (t) = 0.754 * cos ( 7.96 t - 69.52)