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Simple harmonic motion/springs question?
A particle of mass 2 kg lies on a smooth horizontal table attached to one end of a light elastic spring whose other end is fixed to a point A on the table. The particle is at rest at E with the spring just taut when it is suddenly given a velocity of 4m/s towards A. The particle comes to instantaneous rest at a point B where AB = 0.8m. If the natural length of the spring is 1m, find its modulus of elasticity. Prove that the particle performs SHM. Find the time taken to travel from E to B.
Thanks!
1 Answer
- CuglierLv 68 years agoFavorite Answer
ok
spring..L=1.00...m=2 kg..v1=4 m/s...x=0.2m
A......080.............B..,(x= 0.2)...E
Mass at rest in E...any F gives v= 4 m/s
v1=4 m/s at.....E.......Ec= 1/2*m*v^2...Ep=0
Checking units up...it is ok
v2=0 ......at.....B...... Ep=1/2kx^2....Ec=0
then.... Ec=Ep......1/2mv1^2=1/2Kx^2
k=800 new/m...now, F=k*x...F=800*(0,2)
F=160 new..... we know...F=m*a
a=160/2=80 m/s^2.......a=(v1-v2)/t...v2=0
t=4/80=1/20 seg... ANSWER
having Ec ...Ep...and F=kx...there is SHM
Cuglier
