Question 62973: THE PEROD OF A SIMPLE PENDULUM IS DIRECTLY PROPORTIONAL TO THE SQUARE ROOT OF ITS LENGTH. IF A PENDULUM HAS A LENGTH OF 6 FEET AND A PERIOD OF 2 SECONDS, TO WHAT LENGTH SHOULD IT BE SHORTENED TO ACHIEVE A 1 SECOND PERIOD?
Answer by uma(370)   (Show Source):
You can put this solution on YOUR website! If p denotes the period, l the length and K the constant of proportionality then,
p = K*l^2
When l = 6 ft and p = 2 seconds then,
2 = K*(6^2)
==> 2 = K* 36
==> 2/36 = K*36/36
==> 1/18 = K
Thus the equation becomes p = (1/18)*l^2
when p = 1,
1 = (1/18) l^2
multiplying both the sides by 18,
1*18 = l^2
==> 18 = l^2
taking square root on both the sides,
sqrt(18) = l
==> 4.2 = l
Thus to achieve a period of 1 second, the length should be (approx) 4.2 ft.
Good Luck1!!
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