A 240 g mass attached to a horizontal spring oscillates at a frequency of 3.10 Hz . At t =0s, the mass is at x= 5.40 cm and has vx =− 19.0 cm/s . Determine: the position at t=5.40s
answer) for position we have,
x=Acos(4pit+).............1)
A=x2+(v2/
2)
=2pif=2*3.14*3.1=19.468rad/s
A=0.0542+(0.192/19.4682)=5.49cm
=cos-1(5.4/5.49)=0.181rad
so using eqn 1
x=5.49 cos(4pi*5.4+0.181)=2.441cm
so the answer is 2.44 cm or 2.441cm.
A 240 g mass attached to a horizontal spring oscillates at a frequency of 3.10 Hz...
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