Question

A mass m is gently placed on the end of a freely hanging spring. The mass then falls 41 cm before it stops and begins to rise.What is the frequency of the oscillation? Express your answer to two significant figures and include the appropriate units.

Answer 1

Frequency is given by:

f = 1/T

T = time period of motion = 2*pi*sqrt (m/k)

f = (1/(2*pi))*sqrt (k/m)

Now Using energy conservation between initial and final position of mass

KEi + PEi = KEf + PEf

KEi = 0 = KEf, since initial and final velocity of mass is zero

PEi = m*g*x

PEf = (1/2)*k*x^2

So,

m*g*x = (1/2)*k*x^2

k/m = 2*g/x

So, frequency will be:

f = (1/(2*pi))*sqrt (2*g/x)

f = (1/(2*pi))*sqrt (2*9.81/0.41)

f = 1.1 Hz

Please Upvote.