Question

A simple harmonic oscillator consists of a block of mass 1.91 kg attached to a spring of spring constant 100 N/m. When t = 1.00 s, the position and velocity of the block are x = 0.124 m and v = 3.417 m/s.

A. What is the amplitude of the oscillations?

     Answer: 0.488 m

B. What was the position of the block at t = 0 s?

C. What was the velocity of the block at t = 0 s?

Answer 1

1) ? = sqrt(k/m)

=sqrt(100 N/m/ 1.91 kg)

w=7.23 rad/sec

x = A*cos(?t + ?) and v = -?A*sin(?t + ? )

So dividing the 2nd eqn by the first gives

v/x = -?*sin(?t + ?)/cos(?t + ?) = -?*tan(?t + ?)

so (?t + ?) = -arctan(v/(x*?)) = - arctan(3.417/(0.124*7.23)) = -1.31 rad

So ? = -1.31 - 7.43*1 = -8.74 rad

Now plugging into the position eqn we have x = A*cos(?t + ?)

=> A = x/cos(?t + ?)

= 0.124/cos(7.23*1 - 8.74)

A= 0.124m

b)At t = 0 then

x = 0.488*cos(0 - 8.74)

x= 0.48m

c) v = -7.23*0.124*sin(0 - 8.74)

=-0.136m/sec