Question

A 1.00 kg glider attached to a spring with a force constant 25.0 N/m oscillates on a horizontal, frictionless air track. At t = 0, the glider is released from rest at x = -2.70 cm. (That is, the spring is compressed by 2.70 cm.)

(a) Find the period of its motion.
s
(b) Find the maximum values of its speed and acceleration.
m/s
m/s²
(c) Find the position, velocity, and acceleration as functions of time (t).
x(t) =  

cm (position)
v(t) =  

cm/s (velocity)
a(t) =  

cm/s² (acceleration)

Answer 1

(a)

w = sqrt(k/m) = sqrt(25/1) = 5 rad/s

T = 2pi*sqrt(m/k)

T = 2pi*sqrt(1/25)

T = 1.257 sec

(b)

amplitude, Xm = 2.70 cm = 0.027 m

Vmax = w*Xm = 5 rad./s x 0.027 m = 0.135 m/s or 13.5 cm/s

amax = w^2 x Xm = 25 x 0.027 = 0.675 m/s^2 or 67.5 cm/s

(c)

X(t) = Xm*cos(wt + $\Phi$ )

X(t) = -2.70*cos(5t)

v(t) = 13.5*sin(5t)

a(t) = 67.5*cos(5t)

(c)
x = - 2.50 cos[ 6t ] cm.
x' = 2.50 (6)sin[ 6t ] cm/sec.
x'' = 90.00 cos[6 t ] cm/sec^2.