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 = -2.80 cm (that is, the spring is compressed by 2.80 cm).

(a) Find the period of the glider's motion.

1.57 s

b)

speed .112 m/s

acceleration   .448 m/s2

(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/s2, and t is in s. Use the following as necessary: t.)

x(t)= v(t)= a(t)=

x(t)= v(t)= a(t)=

Explanation / Answer

mass of glider, m = 1 kg
force constant of spring, k = 16 N/m
at t = 0, v = 0 and x = -2.8 cm

a. let the equation of motion of the glider be
   x = Asin(wt + phi)
   then v = Awcos(wt + phi)
   at t = 0
   -2.8 = Asin(phi)
   0 = Awcos(phi)

   also, w = sqroot(k/m) = sqroot(16/1) = 4
   so, phi = -90
   A = 2.8 cm

   so, x = 2.8*sin(4t - 90)

   time pertiod = T = 2*pi/w = 2*pi/4 = 1.57 s

b. speed , maximum = Aw = 2.8*4 = 11.2 cm/s = 0.112 m/s
c. acceelration, max = Aw^2 = 0.448 m/s/s
d. position x = 2.8*sin(4t - 90) ( in cm)
   velocity v = 2.8*4cos(4t - 90) = 11.2COS(4T - 90) ( in cm/s)
   acceleration = -kx/m = -44.8sin(4t - 90) (in cm/s/s)

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