Question

The period of oscillation of a simple pendulum increases by 20% when the length of the pendulum is increased by 44 cm. Find its 

(i) initial length 

(ii) initial period of oscillation at a place where g is 9.8 m/s² .

Answer 1

Let T and L be the initial period and length of the pendulum. Let T₁ and L₁be the final period and length. 

Data : T₁ = T + 0.2 T = 1.2 T, L₁ = L + 0.44 m

\(\therefore \) T = 2\(\pi\)\(\sqrt{\frac{L}{g}}\),

T₁ =  2\(\pi\)\(\sqrt{\frac{L_1}{g}}\)

\(\therefore \) \(\frac{T}{T_1}\) = \(\sqrt{\frac{L}{L_1}}\)

\(\therefore \) \(\frac{1}{1.2}\) = \(\sqrt{\frac{L}{L+0.44}}\)

Squaring and cross-multiplying, we get, 

L + 0.44 = 1.44 L 

∴ 0. 44 L = 0.44

∴ L = \(\frac{0.44}{0.44}\) = 1 m

∴ T = 2π \(\sqrt{\frac{L}{g}}\) = 2 × 3.142 × \(\sqrt{\frac{1}{9.8}}\)

= 2.007 s