# Spring oscillation Problem

• bpw91284
In summary, the frequency of the oscillation can be calculated by using the given amplitude and solving for f using the equation f=[1/(2pi)]*[k/m]^0.5, where k is equal to 2mg divided by the amplitude. This can be derived from the equations for potential and kinetic energy, as well as the equation F=kx.

## 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.

Well who could resist such a spring question on the vernal equinox? You have posted many useful eqns, do you have any others that relate the above to periodic motion. In other words, the conditions given will give rise to a pendulum motion, but w/o differential eqns experience or a plug-in formula, its difficult to solve.

from the data given, one can conjecture at the end of the spring bob:(energy conservation)

1/2Ky^2=mgy where Y=.36m hence, k=2mg/y. So now we have K. Most problems of thsi sort have k and m in a radical, any help yet?

denverdoc said:
Well who could resist such a spring question on the vernal equinox? You have posted many useful eqns, do you have any others that relate the above to periodic motion. In other words, the conditions given will give rise to a pendulum motion, but w/o differential eqns experience or a plug-in formula, its difficult to solve.

from the data given, one can conjecture at the end of the spring bob:(energy conservation)

1/2Ky^2=mgy where Y=.36m hence, k=2mg/y. So now we have K. Most problems of thsi sort have k and m in a radical, any help yet?

Don't have time now, but I'll look at it later. It is an introductory physics class so I know no diffy q is needed.

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

F=kx
mg=kx
k/m=g/x, were x is the amplitude which is known, substitute into above "f" equation and solve. That valid?