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=100981=9.81 N  F=k d  k=F/d=9.81/0.015=654  f=2π1 mk=2 3.14161 1654=4.0701 Hz

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