Lowest natural frequency of vibration

Two otherwise identical ropes differ in length
L by a factor of 2. If the two ropes are fixed at their
ends, the lowest natural frequency of vibration of the shorter rope is:

1. Twice that of the longer rope.
2. sqrt2 larger than that of the longer rope.
3. The same as that of the longer rope.
4. 1/sqrt2 smaller than that of the longer rope.
5. Half that of the longer rope.

Assuming they have the same tension.

The lowest frequency (fundamental) occurs when they each have one node at each end (node-node distance = wavelength/2).

So the shorter rope has half the wavelength of the longer one
Speeds, v, of waves on the ropes are the same (identical mass per unit length, same tension)
v = lambda x f
f = v/lambda
So the shorter rope has double the frequency of the longer one. Answer 1.