Question

the time periods of two simple pendulums at a place are in ratio 2:1. what will be the ratio pf their length

Answer 1

T=2π√[l/g]
we can se T² is directly proportional to length
so L1/L2=(T1/T2)²
T1/T2 is 2:1
so L1/L2=4/1

Answer 2

The ratio of their length is 4 : 1

Given:

The time period of two simple pendulum are in the ratio of 2 : 1  

Solution:

Time period of a simple pendulum is calculated by the formula,

$T=\frac{2 \pi \sqrt{l}}{g}$

Let us assume that the time period for first pendulum be T1 and for second pendulum be T2,

Thereby we have,

$T_{1}=\frac{2 \pi \sqrt{l_{1}}}{g} \rightarrow (1)$

$T_{2}=\frac{2 \pi \sqrt{l_{2}}}{g} \rightarrow (2)$

Dividing equation (1) and (2), we get,

$\frac{T_{1}}{T_{2}}=\frac{\sqrt{l_{1}}}{\sqrt{l_{2}}}$

On squaring both sides, we get,

$\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{l_{1}}{l_{2}}$

$\frac{l_{1}}{l_{2}}=\frac{2^{2}}{1^{2}}$

$\frac{l_{1}}{l_{2}}=\frac{4}{1}$

Therefore, the length of the two simple pendulum be 4 : 1