GRE GR 1777, Problem 003 Solution

Physics GRE GR 1777

3. Classical Mechanics (Oscillations)

Solution) Angular frequency : $\omega = \frac{2\pi}{T}$

Period : $T = \frac{2\pi}{\omega}$

In the simple harmonic pendulum, $\omega = \sqrt{\frac{g}{L}}$
where L is the length of the pendulum.

Using above equations, $T = 2\pi \sqrt{\frac{L}{g}}$ , then we can see that $L \propto T^2$ .

Since $T_{B} = 2 T_{A}$ , $T_{A} : T_{B} = 1 : 2$ .
Then $L_{A} : L_{B} = T^2_{A} : T^2_{B} = 1 : 4$ , therefore $L_{B} = 4 L_{A}$ .

The answer is (E).