Asked by Andrea
A block oscillates on a spring and passes its equilibrium position with a speed of .157m/s. It's kinetic energy is zero when the block is at the distance of .1m from equilibrium. Assume no friction between the block and the table. What is the period of its oscillation?
v=.157m/s
KE =0J when x=.1m
I know that T=2pi sqrt(m/k)
yet I am not given mass of the k(the spring constant)
PE=mgh
PE(max)= m(9.81m/s^2)(.1m)
but again i'm not given m.
How would I solve this problem?
v=.157m/s
KE =0J when x=.1m
I know that T=2pi sqrt(m/k)
yet I am not given mass of the k(the spring constant)
PE=mgh
PE(max)= m(9.81m/s^2)(.1m)
but again i'm not given m.
How would I solve this problem?
Answers
Answered by
bobpursley
energy in spring at end=1/2 k (.1)^2
set that equal to ke at equilibrium.
1/2 mv^2=1/2 k (.1)^2
solve for m/k
set that equal to ke at equilibrium.
1/2 mv^2=1/2 k (.1)^2
solve for m/k
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