A simple pendulum has a period of 17 0s When th... - JAMB Physics 2000 Question
A simple pendulum has a period of 17.0s. When the length is shorten by 1.5m, its period is 8.5s. Calculate the original length of the pendulum
A
4.0 m
B
3.0 m
C
2.0 m
D
1.5 m
correct option: c
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
√ 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
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