Question

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 . At the block has velocity -4.00 and displacement +0.200 .


Part A
Find (a) the amplitude and (b) the phase angle.
=




Part B
φ=




Part C
Write an equation for the position as a function of time.
Assume in meters and in seconds.
=

Answer 1

V^2 = k/m.(A^2-Y^2)
16 = 300/2 (A^2 - 0.2^2)
16 = 150 (A^2 - 0.2^2)
16 = 150A^2 - 6.00
22 = 150A^2
A=0.38 m

a= W^2.A
a= 300/2 x 0.38
a=57.4 m/s^2

F=k.A
F=300 x 0.35
F=114.9Newton