Asked by Sammy
A 7 kg mass hangs on a 10 meter long weightless cord. To the nearest cm, what should the new length of the pendulum be in order that the new frequency be 1.8 times its current value?
The correct answer is 309 but I got 250.
Heres what I did..
(1/2pi)(sqrt g/L2) = (2/2pi)(sqrt g/L1)
1/L2 = 4/L1
4L2 = L1
L2 = L1/4
So 10/4 times 100cm/1m = 225...
The correct answer is 309 but I got 250.
Heres what I did..
(1/2pi)(sqrt g/L2) = (2/2pi)(sqrt g/L1)
1/L2 = 4/L1
4L2 = L1
L2 = L1/4
So 10/4 times 100cm/1m = 225...
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