After a 10 pounds weight is attached to a 5 foot spring, the spring measures 7 f

Question

After a 10 pounds weight is attached to a 5 foot spring, the spring measures 7 feet long. The 10 pound weight is removed and replaced with an 8 pound weight, and the entire system placed in a medium offering resistance numerically equal to the instantaneous velocity...
A) find the equation of motion if the weight is released 1/2 foot below the equilibrium position with a downward velocity of 1 ft/s
B)What are the amplitude and period of the motion? how many complete vibrations the weight completed at the end of 2p seconds?
C)Find the times at which the weight passes through the equilibrium position heading downward?

Explanation / Answer

hey start with the basic equations 10/32=(5/16)slug. 2=k/m , then find your general solution.

x(t)=c1cos(t)+c2sin(t)

then you can solve for c1 and c2.    try using something like    x(0)=+distance since below equalibrium

play with it                                                                              xI(0)=velocity

Amplitude=c12+c22

tan=c1/c2

x(t)=Asin(t+)     (equation of motion) hint: sin(x+y)=sinxcosy+sinycosx

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