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

c. √2T

Step-by-step explanation:

The period of a simple pendulum is given by;

`T = 2\pi \sqrt{(L)/(g) } \\\\(T)/(2\pi) = \sqrt{(L)/(g)} \\\\(T^2)/(4\pi^2) = (L)/(g)\\\\(g)/(4\pi^2) = (L)/(T^2)\\\\ (L_1)/(T_1^2)= (L_2)/(T_2^2)\\\\T_2^2 = (L_2T_1^2)/(L_1)\\\\L_2 = 2L_1\\\\ T_2^2 = (2L_1T_1^2)/(L_1)\\\\ T_2^2 =2T_1^2\\\\T_2 = √(2T_1^2)\\\\T_2 = T_1√(2)`

Thus, the the new period will be √2T