Question

A 0.11 kg mass suspended on a spring is pulled to 6.9 cm below its equilibrium position and released. When the mass passes through the equilibrium position, it has a speed of 0.28 m/s. What is the speed of the mass when it is 4.1 cm from the equilibrium position?

Answer 1

x=A•cos(ωt),
A = 0.069 m
v =A•ω•sin(ωt),
at equilibrium position v = v(max)= A• ω = 0.28 m/s
ω = 0.28 /A=0.28/0.069 =4.058 rad/s.
x=A•cos(ωt),
cos(ωt) = x/A = 0.041/0.069 =0.594,
sin (ωt) = sqrt(1 – (x/A)^2)=0.804,
v1 =A•ω•sin(ωt) =0.069•4.058•0.804 =0.225m/s.