Posted by **Sammy** on Saturday, July 13, 2013 at 2:29pm.

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...

- College Physics -
**Sammy**, Saturday, July 13, 2013 at 2:41pm
Sorry, I meant 250 not 225. Either way its still not the correct answer..

- College Physics -
**Elena**, Saturday, July 13, 2013 at 4:31pm
ω₁= √(g/L₁)

ω₂=√(g/L₂)

1.8 ω₁=ω₂

1.8√(g/L₁)=√(g/L₂)

1.8²g/L₁=g/L₂

L₂=L₁/3.24=10/3.24 =

=3.086 ≈3.19 m = 309 cm

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