Question

2.0 kg weight is on simple, horizontally oscillation (in a spring) with the time 1.8 s.
The distance between the outmost dots in the oscillation is 0.16 m.

a) What is the angular speed (w) of the oscillation ?
b) What is the force constant (k) of the spring?
c) What is the total energy of the oscillation ?
d) What is the speed of the weight when it goes through the central position of the oscillation?
e) What is the speed when lateralisation is 5 cm ?

I found out out part a) and b) :

a) w=2pi/T
w= 2pi/1.8= 3.49 s^-1

b) k=(2 pi)^2/T^2
k=(2pi)^2 x 0.80 / 1.8^2 = 24 N/m

Can someone please explain for me how to do part c)-e) ?

Given answers are: c) 0.077 J; d) 0.28 m/s; e) 0.22 m/s

Answer 1

b)
w = 2 pi/T = sqrt (k/m)

(2 pi)^2 /T^2 = k/m

k = 2 (2 pi)^2/ 1.8^2
= 24.4 n/m
(I do not know how you got the same answer) !!!!

c) when it is stopped at x = .08 meters, all the energy is potential (1/2) k x^2

(1/2)(24.4) (.08)^2 = .0781 Joules

d) Ke = .0781 Joules = (1/2)mv^2
.0781 = v^2
v = .279 m/s

e) if x = .05
(1/2) k x^2 = (1/2)(24.4)(.05)^2
= .035 Joules
so Ke = .0781 - .035 = .0476 Joules
so (1/2) m v^2 = .0746
v = .218 m/s