A simple pendulum has a period of 17 0s When th... - JAMB Physics 2000 Question

For a simple pendulum, the period T = 2π

√  ¹/_g 

Since g, 2 and π are constant, => T ∝ √l

∴ T = K√l

Let the original length = l

thus for the length l, period T = 17s

∴ 17 = K√1 .................. [1]

Again, for the new length [1 - 1.5]

the period T = 8.5s
∴ 8.5 = k

√ [1-1.5] ...............[2]

thus dividing eqn [1] by eqn[2] we have
¹⁷/_8.5 = √(l/l-1.5)

∴2 = √(l/l-1.5)

=>4 = (l/l-1.5)

∴l = 4[l-1.5]

= 4l - 6.0

∴4l - l = 6.0

3l = 6.0

l = 2.0m