A 100-g block hangs from a spring with k = 5.3 N/m. At t = 0 s, the block is 20.0 cm below the equilibrium position and moving upward with a speed of 194 cm/s. What is the block's speed when the displacement from equilibrium is 31.5 cm?

Answer 1

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•6.76/194=0.7
ωt =arctanωt =0.61 rad,
sinωt = 0.571
A=x/sinωt = 20/0.571=35 cm.
For the second case:
x1=A•sinωt1,
sinωt1 =x1/A=31.5/35 =0.9.
cos ωt1=sqrt(1-sin²ωt1)=0.44.
v1= =A•ω•cosωt1=
=35•6.76•0.44=104 cm/s.

Answer 2

K is 5.3 N/m. Why are you using 4.6 N/m?

Answer 3

This is the problem similar to that I've solved for another post. Substitute your data and calculate yourself

Answer 4

Check my calculations.
ω=sqrt(k/m)=sqrt(5.3/0.1) = 7.28rad/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•7.28/194=0.75
ωt =arctanωt =0.64 rad,
sinωt = 0.6
A=x/sinωt = 20/0.571=33.3 cm.
For the second case:
x1=A•sinωt1,
sinωt1 =x1/A=31.5/33.3 =0.95.
cos ωt1=sqrt(1-sin²ωt1)=0.33.
v1= =A•ω•cosωt1=
=33.3•7.28•0.33=80 cm/s.

Answer 5

I got a similar answer of 79.0 cm/s using the conservation of energy equation.

U(i) + K(i) = U(f) + U(f)

Thank you for the help. And sorry for asking this question. I didn't see the similar question when I searched mine in the search bar.