The time periods of two simple pendulums at a place are in the ratio 2:1. What will be the ration of their lengths?

Open in App

Solution

We know that time period of pendulum is:
T = 2 $π \sqrt{\frac{l}{g}}$
As per the above relation,, time period is proportional to the square root of the length.
Let T (1) and T (2) be the time period and L (1) and L (2) be the length of the pendulum respectively, accordingly;
T (1) : T (2) = $\sqrt{} L (1) : \sqrt{} L (2)$
Or
$L 1 : L 2$ = $(T l)^{2}$ : $(T 2)^{2}$
$L 1 : L 2$ = $(2)^{2}$ : $(1)^{2}$
= $4 : 1$

Suggest Corrections

Similar questions

Q. The time period of two pendulum at a place are in ratio 2:1 . What will be the ratio of the length

Q. What is the ratio the time periods of two pendulums at a given place, when their lengths are

64 c m

and

256 c m

respectively.

Q. Two simple pendulums A and B have equal length, but their bobs weigh 50 g and 100 g respectively, the ratio of their time periods is.

Q. Two simple pendulums A and B have equal lengths, but their bobs weigh 50 gf and 100 gf respectively. What would be the ratio of their time periods? Give reason for your answer.

The time periods of two simple pendulums of different lengths are the same on two different planets. If the lengths of the two pendulums are in the ratio of $1 : 9$ , then the ratio of the accelerations due to gravity on the two planets, most likely, is :

Join BYJU'S Learning Program