Question

1. Down at the ice skating rink you watch a customer try out the new exercise machine consisting of a large spring allowing the skater to oscillate back and forth. The skater (mass-64kg) is stretched 0.84m from equilibrium with a force 250N and remains at rest. At 0.84m from equilibrium, the skater is then pushed toward the equilibrium position. At a distance of 0.22m (from equilibrium), the skater has a speed of 2.4m/s. a) Starting with Newton's second law, show that this system obeys the simple harmonic oscillator differential equation. Include a free body diagram with positive axis labeled. What is the general solution of this equation? b) Write down the equation describing the skater's position as a function of time (assume t-0 at 0.22m and motion is in the negative x-direction). Your solution should have numeric values for the variables. your answer in b) above. Be sure to include numbers on each axis. d) How much time elapses for the skater to travel the distance from the starting position, x-0.22m, directly to equilibrium (x-0)? e) What is the skater's speed at a time of 1.2s?

Answer 1

A = 0.84m; F = 250N ; x = 0.22m and v = 2.4m/s ; m = 64kg;

a = 250/64 = 3.91 m/s^2

but a = w^2A

w = 2.16 rad/s

k/m = w^2

k = mw^2 = 64x2.16^2 = 299 N/m

ma + kx = 0

64a + 299x = 0 ... general solution

x = Asin(wt+@)

0.22= 0.84 sin(2.16 x0 + @)

@ = 15.18 deg = 0.265 rad

x = 0.84sin(2.16t + 0.265)

5 4 3 0 0 (u) uousod

0 = 0.84sin(2.16t+0.265)

2 x 3.14 = 2.16t + 0.265

t = 2.785 s