A partridge of mass 5.01 kg is suspended from a pear tree by an ideal spring of

Question

A partridge of mass 5.01 kg is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.21 s.

I've already found:
speed through equilibrium position v=.149 m/s
acceleration at .05m above equilibrium position a= -.111 m/s^2
time required from .05 below equilibrium position to .05 above t= .702 s

The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

Explanation / Answer

= 2 pi/T

k = m^2

kx = mg

So

x=mg/k = 5.01*9.8/(5.01 * (2 pi /4.21)^2 )= 4.4 m 

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