Question

The time periods of two simple pendulums are in the ratio 2:1,
(i) Compare their lengths.
(ii) Give a reason for your answer

Answer 1

Answer:

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}T=g2πl

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)T1=g2πl1→(1)

T_{2}=\frac{2 \pi \sqrt{l_{2}}}{g} \rightarrow (2)T2=g2πl2→(2)

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

\frac{T_{1}}{T_{2}}=\frac{\sqrt{l_{1}}}{\sqrt{l_{2}}}T2T1=l2l1

On squaring both sides, we get,

\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{l_{1}}{l_{2}}T22T12=l2l1

\frac{l_{1}}{l_{2}}=\frac{2^{2}}{1^{2}}l2l1=1222

\frac{l_{1}}{l_{2}}=\frac{4}{1}l2l1=14

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

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