Frequency of oscillation

A weight of 1 kg was suspended from the spring and the spring was extended by 1.5 cm. Determine the frequency of the natural oscillation of the resulting oscillator.

Correct answer:

f =  4.0701 Hz

Step-by-step explanation:

m=1 kg d=1.5 cm m=1.5/100  m=0.015 m g=9.81 m/s2  F=m g=1 9.81=981100=9.81 N  F=k d  k=F/d=9.81/0.015=654  f=12π km=12 3.1416 6541=4.0701 Hz



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