A 1.00 kg glider attached to a spring with a force constant 16.0 N/m oscillates

Question

A 1.00 kg glider attached to a spring with a force constant 16.0 N/m oscillates on a frictionless, horizontal air track. At t-0, the glider is released from rest at x =-3.10 cm that is, the spring is compressed by 3.10 cm). (a) Find the period of the glider's motion Domestic Transfer Application your applicant ID is 1275458 https://admissions.msu.edu/application/Application.asp acceleration (c) Find the position, velocity, and acceleration as functions of time. (Where position is in m, velocity is in m/s, acceleration is in m/s, and t is in s. Use the following as necessary: t.) x(t) = r(t) = att)

Explanation / Answer

Given,

K = 16 M/m ; x = -3.1 cm ; m = 1

a)We know that

w = sqrt(k/m)

w = sqrt (16/1) = 4

w = 2 pi/T => T = 2 pi/w

T = 2 x 3.14/4 = 1.57 s

Hence, T = 1.57 s

b)a = A w^2

a = 0.031 x 4^2 = 0.496 m/s^2

Hence, a = 0.496 m/s^2

c)x(t) = A cos(wt)

x(t) = -0.031 cos(4t)

v(t) = A w sin(wt) = -0.031 x 4 sin(4t)

v(t) = -0.124 sin(4t)

a(t) = A w^2 cos(wt) = -0.031 x 4^2 cos(4t)

a(t) = -0.496 cos(4t)

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