ω=sqrt(k/m)=sqrt(4.6/0.1) = 6.76rad/s.
x=A•sinωt
v=dx/dt=A•ω•cosωt
Divide the first equation by the second and obtain
x/v= A•sinωt/ A•ω•cosωt= ω•tanωt,
tanωt = x•ω/v=20•6.78/214=0.634.
ωt =arctanωt =0.565 rad,
sinωt = 0.535
A=x/sinωt = 20/0.535=37.35 cm.
For the second case:
x1=A•sinωt1,
sinωt1 =x1/A=32.1/37.35 =0.86.
cos ωt1=sqrt(1-sin²ωt1)=0.511.
v1= =A•ω•cosωt1=
=37.35•6.78•0.511=129.5≈130 cm/s.
A 100-g block hangs from a spring with k = 4.6 N/m. At t = 0 s, the block is 20.0 cm below the equilibrium position and moving upward with a speed of 214 cm/s. What is the block's speed when the displacement from equilibrium is 32.1 cm?
The answer is supposed to be 130 cm/s; however, I'm not sure how to get this answer. I've been trying to use a conservation of energy equation, but I'm not getting the correct answer. Any help is greatly appreciated.
2 answers
I'm a little lost.
How is wt = arctan(wt) = 0.565 rad & sin(wt) = 0.535 rad?
How is wt = arctan(wt) = 0.565 rad & sin(wt) = 0.535 rad?