A mass weighing 10 pounds stretches a spring 1/4 foot. This mass is removed and replaced with a mass of 1.6 slugs, which is initially released from a point 1/3 foot above the equilibrium position with a downward velocity of 5/4 ft/s.
At what time does the mass attain a displacement below the equilibrium position numerically equal to 1/2 the amplitude?
my position function is
I have tried to put it into the form
and I am getting
The time I am getting is initially, but I don't know if this is right. I'm also having trouble writing the answer as a series. Can anyone help me? Thanks
I thought I had to convert the mass into slugs by , which gives the force as slugs for Hooke's law. This gives . Then I found m for the weight of slugs >>> where This gives . Is this incorrect?
Shouldn't your equation be ?
I did it again without rounding off as much and got:
the time I got:
Am I on the wrong track with this? Thanks a lot!