4. A simple harmonic oscillator consists of a 0.5-kg block attached to a spring

Question

4. A simple harmonic oscillator consists of a 0.5-kg block attached to a spring and vibrates about an equilibrium position at x- 0. The force constant of the spring is 32 N/m and the block oscillates on a horizontal frictionless surface described by this general equation for its displacement as a function of time, x(1) (15 cm) cos (or + ?) , with angular speed (c) and phase constant ((p) of this oscillation. At timet-0.2 s, the block is at a displacement x 5.0 cm from its equilibrium position. (a) Find the angular speed (o) and phase constant (p) of this oscillation (b) Determine the instantaneous velocity of this block at 1.5 s? (c) Using energy methods or otherwise, find the speed of this block at the instant the block is at a displacement x 0.10 m from its equilibrium position, after being released?

Explanation / Answer

given

m = 0.5 kg
k = 32 N/m
A = 15 cm

a) w = sqrt(k/m)

= sqrt(32/0.5)

= 8 rad/s

x(t) = 15*cos(8*t + phi)

at t = 0 , x = -5.0 cm

-5 = 15*cos(0 + phi)

cos(phi) = -1/3

phi = cos^-1(-1/3)

phi = 1.91 radians

b) so, x(t) = 15*cos(8*t + 1.91)

v = dx/dt

= 15*(-sin(8*t + 1.91)*8)

= -120*sin(8*t + 1.91)

at t = 1.5 s

v = -120*sin(8*1.5 + 1.91)

= -117 cm/s or -1.17 m/s

c)

(1/2)*m*v^2 + (1/2)*k*x^2 = (1/2)*k*A^2

(1/2)*0.5*v^2 + (1/2)*32*0.1^2 = (1/2)*32*0.15^2

==> v = 0.894 m/s

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