A simple pendulum of length l 1 has periodic time 2.4 s. Another simple pendulum of length l 2

Question

A simple pendulum of length l 1 ​ has periodic time 2.4 s. Another simple pendulum of length l 2 ​ < l 1 ​ , has periodic time 1.8 s. The periodic time of simple pendulum of length ( l 1 ​ − l 2 ​ ) is nearly

TestHub · Accepted Answer

The time period of a simple pendulum is given by $T = 2 π \frac{l}{g}$ .

Squaring both sides, we get $T^{2} = \frac{4 π ^{2} l}{g}$ , which implies $l = \frac{g T ^{2}}{4 π ^{2}}$ .

For the first pendulum, $l_{1} = \frac{g T _{1}^{2}}{4 π ^{2}}$ .

For the second pendulum, $l_{2} = \frac{g T _{2}^{2}}{4 π ^{2}}$ .

For the pendulum of length $l_{1} - l_{2}$ , the time period $T$ is given by:

$T^{2} = \frac{4 π ^{2}}{g} (l_{1} - l_{2}) = \frac{4 π ^{2}}{g} (\frac{g T _{1}^{2}}{4 π ^{2}} - \frac{g T _{2}^{2}}{4 π ^{2}}) = T_{1}^{2} - T_{2}^{2}$

Substituting the given values $T_{1} = 2.4$ s and $T_{2} = 1.8$ s:

$T^{2} = (2.4)^{2} - (1.8)^{2} = 5.76 - 3.24 = 2.52$

$T = 2.52 \approx 1.587$ s

This is nearly $1.6$ s.

Answer: $1.6$ s

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