Asked by Kelly O'Reilly
A 2kg mass is attached to a spring of spring constant 20N/m. At t=0 the spring is at its relaxed position and the block is moving with a velocity of 4m/s to the right. Assume the positive direction is to the right. What is the speed of the mass when it is 1 meter from the equilibrium position?
A) 1.6 m/s
B) 2.4 m/s
C) 2.5 m/s
D) 3.1 m/s
E) 3.4 m/s
A) 1.6 m/s
B) 2.4 m/s
C) 2.5 m/s
D) 3.1 m/s
E) 3.4 m/s
Answers
Answered by
Elena
ω=sqrt(k/m) = sqrt(20/2) = 3.16 rad/s
x=Asinωt
v=Aωcosωt
t=0 => cosωt=1
A=v/ω =4/3.16=1.265 m
x=Asinωt
sinωt =x/A=1/1.265 =0.79
cos ωt =sqrt{1-sin²ωt} = 0.612
v₁=Aωcosωt =1.265•3.16•0.612 =2.4464 m
x=Asinωt
v=Aωcosωt
t=0 => cosωt=1
A=v/ω =4/3.16=1.265 m
x=Asinωt
sinωt =x/A=1/1.265 =0.79
cos ωt =sqrt{1-sin²ωt} = 0.612
v₁=Aωcosωt =1.265•3.16•0.612 =2.4464 m
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