Question

A certain frictionless simple pendulum having a length L and mass M swings with period T. If both L and M are doubled, what is the new period?
a. T
b. 2T
c. √2T
d. T/2
e. T/4

Answer 1

Answer:

c. √2T

Explanation:

The period of a simple pendulum is given by;

$T = 2\pi \sqrt{\frac{L}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}} \\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\\frac{g}{4\pi^2} = \frac{L}{T^2}\\\\ \frac{L_1}{T_1^2}= \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{L_2T_1^2}{L_1}\\\\L_2 = 2L_1\\\\ T_2^2 = \frac{2L_1T_1^2}{L_1}\\\\ T_2^2 =2T_1^2\\\\T_2 = \sqrt{2T_1^2}\\\\T_2 = T_1\sqrt{2}$

Thus, the the new period will be √2T