A 5.0 kg block hangs from a spring with spring constant 2000N/m. The block is pulled down 5 cm from equilibrium position and given an initial velocity of 1.0 m/s back toward equilibrium. What are the
c.) total mechanical energy of the motion
oscillatory motion x(t) = Acos(ωt + ø0)
f = ω/(2∏)
ω = √(k/m)
The Attempt at a Solution
frequency found by ω = √(k/m)
ω = √(2000/5) = 20
f = 20/(2∏) = 10/∏
which I believe was correct.
The only way to find the amplitude is to relate it to the velocity, correct? Here I can't simply use .5kA2 = .5m(Vmax)2 because at Vmax there should also be some gravitational potential. Used the motion equation and it's derivative:
x(t) = Acos(ωt + ø0)
v(t) = -ωAsin(ωt + ø0)
x(0) = .05 = A cos(ø0)
=>ø0 = cos-1(.05/A)
edit: here I made an error v(0) should be -20Asin(ø0)
v(0) = -1 = -20A/sin(ø0)
-1 = -20Asin(cos-1(.05/A))
=>sin-1(1/(20A)) = cos-1(.05/A)
which is the same thing as saying (.05)2 + (.05)2 = A2 right? If so, A = .0707, which I can't verify at the moment but seems reasonable.
Where I get stuck is here:
The energy should be entirely kinetic at the bottom of the stretch given by .5k(A + Δx)2 where Δx was the initial stretch of the spring caused by gravity before it was pulled. So I figured I could set it equal to the initial condition when the spring is displaced .05 m from equilibrium and the initial velocity is 1.0 m/s:
.5k(A + Δx)2 = .5k(.05 + Δx)2 + mg(A-.05) + .5mV02
Then I solved for Δx, in order to plug it back into .5k(A + Δx)2 and I won't bore you with all that here but I've done it twice now and Δx comes out -.1549, which doesn't seem right. Plugging it into the equation negative or positive comes out as 26.6 and 223.6 respectively, and I believe the answer was supposed to be 5 as total mechanical energy.
After doing it again with the correct amplitude I got 8.94, a lot closer to 5.0 but still off...
If anyone sees my errors here I'd be super grateful!