Asked by Shandy
a vertical spring with spring stiffness constant 305 n/m oscillates with an amplitude of 28.0 cm when 0.260 kg hangs from it. the mass passes through the equilibrium point (y=0) with positive velocity at t=0. (a) what equation describes this motion as a function of time? (b) at what times will the spring be longest and shortest?
Answers
Answered by
Elena
(a)
ω=sqrt(k/m) = sqrt(305/0.26) = 34.2 rad/s
y=Asin ωt = >
y=0.28sin34.2t (meters)
(b) y=A (longest)
A= Asin ωt
sin ωt = 1
ωt= π
t= π/ω=3.14/34.2= 0.092 s.
T=2π/ ω = 2•3.14/34.2 = 0.183 s
The shortest spring after T/2 =>
t₁ =t+ (T/2) = 0.092 +(0.183/2) =
=0.183 s
ω=sqrt(k/m) = sqrt(305/0.26) = 34.2 rad/s
y=Asin ωt = >
y=0.28sin34.2t (meters)
(b) y=A (longest)
A= Asin ωt
sin ωt = 1
ωt= π
t= π/ω=3.14/34.2= 0.092 s.
T=2π/ ω = 2•3.14/34.2 = 0.183 s
The shortest spring after T/2 =>
t₁ =t+ (T/2) = 0.092 +(0.183/2) =
=0.183 s
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