Question

An object attached to a spring with a constant of 200 N/m vibrates at 18 Hz. If the stiffness of the spring is doubled, how would the oscillations change?

Answer 1

Explanation:

k = 200 n/m

f = 18 Hz

K = 2k

¹

t = 2π √m/k

f = 1/t

f = 1/2π √k/m ---------1

¹

f = 1/2π √k/m ----------2

² ¹

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divide 1 & 2

f / f = √k/k

¹ ² ¹

18/f = √1/2

²

Squaring both these sides

(18)²/f = 1/2

²

324/f = 1/2

²

f = 324×2

²

f = 648 Hz

²