How does period compar- Don't understand solution

1. May 12, 2014

eidyllion

1. The problem statement, all variables and given/known data
Suppose that you make careful observations of a pendulum on the surface of the Earth. You then move to Mars, where g is about 40% of the value on the surface of the Earth. How does the period of the pendulum on Mars compare with the period of the pendulum on Earth The period on Mars is _____ the period on Earth.

The answer is B) Larger by a factor of 1/sqrt(0.4) than

2. Relevant equations
Angular frequency: ω= sqrt(g/L) = 2∏/T
This was achieved because I know that ω=2π/T and that T=2π*sqrt(L/g)

3. The attempt at a solution
My teacher's solution says:
T=2∏*sqrt(4g)
T is g^(-1/2)
so as g decreases→T increases

I don't understand his solution. Where does sqrt(4g) come from and why does that mean T is g^(-1/2)

2. May 12, 2014

Nathanael

Everything is the same between the Earth equation and the Mars equations except for one thing, g is replaced with 0.4g

If the period were proportional to g, it would be 0.4 times longer on Mars as on Earth. (Which would actually be shorter since 0.4 is less than 1)

But as you can see by that equation, it is actually proportional to the inverse sqrt(g).

So it becomes longer by a factor of the inverse of the sqrt(0.4)

(Sorry I can't think of how to make it much clearer than this, but hopefully this helps.)

EDIT:
"T is g^(-1/2)
so as g decreases→T increases"

"T is (proportional to) g^(-1/2)
so as g decreases→T increases"

3. May 12, 2014

dauto

There is no "T=2∏*sqrt(4g)". That's just a typo. Skip that line.