Question

On suspending a 1.0 kg mass by a massless spring, there is increase of 1.0 cm in its length. Find: (i) Force constant of the spring (ii) Time period of the vibration of the spring.

Answer 1

m = 1.0 kg; l = 1.0 cm = 0.01 m; g = 9.8 m s^-2

(i) ∵ F = ky

or mg = ky ⇒ k  = \(\frac{mg}{y} = \frac{1.0\times9.8}{0.01}\)

or k = 980 N m^-1

(ii) ∵ T = 2π\(\frac{m}{k}\)

\(\therefore\ T = 2\pi\sqrt{\frac{1}{980}} = \frac{2\times3.14}{31.3049}\)

\( = \frac{6.28}{31.3049}\) 

\(T = 0.2006\ s\)