Post a New Question

college physic SHM

posted by on .

A 1.25kg mass stretches a vertical spring 0.215m.If the spring is stretches an additional 0.130m and released,
(a)how long does it take to reach the equilibrium position again?
(b)Write an equation giving its position y as a function of time t.
(c)What will be its maximum speed and maximum acceleration?

  • college physic SHM - ,

    The spring constant is
    k = Mg/0.215m = 57.0 N/m

    The frequency of oscillation is
    f = [1/(2 pi)]sqrt(k/m) = 1.075 Hz
    The angular frequency is
    w = sqrt(k/m) = 6.75 rad/s

    The period is 1/f = 0.93 seconds

    (a) 1/4 period = 0.2325 s is the time needed to return to the equilibrium position

    (b) If y is measured from the equilibrium position

    y = 0.13 cos wt = 0.14 cos6.75t

    (c) Max speed = w*(amplitude)= 0.13 w
    Max acceleration = w^2*(amplitude)

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question