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...
2 answers
Sorry, I meant 250 not 225. Either way its still not the correct answer..
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
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
0 answers