Question

5 kg mass is hung on the end of a helical spring and is pulled down and let go so as to vibrate vertically. The mass completes 100 vibrations in 60 seconds. Calculate the spring constant factor. *​

Answer 1

vidikhkgyujgyikjggy TT the

Answer 2

Given info : 5 kg mass is hung on the end of a helical spring and is pulled down and let go so as to vibrate vertically. The mass completes 100 vibrations in 60 seconds.

To find : calculate the spring constant factor.

solution : at equilibrium,

centripetal force is balanced by spring force.

i.e., mω²x = Kx

⇒mω² = K....(1)

here , mass completes 100 vibrations in 60 seconds.

so, ω = 100/60 rev/s = 100 × 2π/60 = 10.47 m/s

from equation (1),

5 kg × (10.47 m/s)² = K

⇒548.1045 = K

Therefore the spring constant factor is 548.1045 N/m.

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