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## Homework Statement

A mass m is gently placed on the end of a freely hanging spring. The mass then falls 36 cm before it stops and begins to rise. What is the frequency of the oscillation?

## Homework Equations

f=[1/(2pi)]*[k/m]^0.5

E=KE+PE

PE_s=0.5kx^2

KE=0.5mv^2

v=rw

## The Attempt at a Solution

So all we start off know is the amplitude is 36cm.

At a peak of oscillation velocity=0 so,

E=PE+KE => KE=0, E=PE

E=0.5*kA^2

At equilibrium point (middle of oscillation velocity=max and PE=0)

E=KE

0.5*kA^2=0.5*mv^2

v_max=wA so,

0.5*kA^2=0.5*m*w^2*A^2, A's and 0.5's cancel out (bad because only value given?)

k=mw^2, w=2(pi)f

k=m[2(pi)f]^2

Solve for f and I just did a proof of f=[1/(2pi)]*[k/m]^0.5 on accident and got no where...help.