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

## Answer This Question

## Related Questions

- physics - A block of mass 4.44 kg lies on a frictionless horizontal surface. The...
- physics - A block of mass M hangs from a rubber cord. The block is supported so ...
- Physics - Two different simple harmonic oscillators have the same natural ...
- Physics - An elastic cord vibrates with a frequency of 2.7 Hz when a mass of 0....
- Physics-Mechanics - horizontal surface. The block is connected by a cord passing...
- Physics - A block of mass M hangs from a rubber cord. The block is supported so ...
- physics - 5. A 2.00-kg block hangs from a rubber cord, being supported so that ...
- Physics - A block of mass 5.53 kg lies on a frictionless horizontal surface. The...
- physics - A block of mass 3.62 kg lies on a frictionless horizontal surface. The...
- Physics 40 - A pendulum bob of mass M1 oscillates with a period of 1.0 s. If the...

More Related Questions