Two simple pendulums of lengths L 1 and L 2 have periodic time T 1 and T 2 respectively ( T 1 T 2 ) . The time period of the pendulum of length ( L 1 − L 2 ) is [( L 1 − L 2 )

Question

Two simple pendulums of lengths L 1 ​ and L 2 ​ have periodic time T 1 ​ and T 2 ​ respectively ( T 1 ​ > T 2 ​ ) . The time period of the pendulum of length ( L 1 ​ − L 2 ​ ) is [( L 1 ​ − L 2 ​ ) > 60 cm ]

TestHub · Accepted Answer

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

Squaring both sides gives $T^{2} = \frac{4 π ^{2} L}{g}$ , 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, $L_{2} = \frac{g T _{2}^{2}}{4 π ^{2}}$ .

The length of the new pendulum is $L^{'} = L_{1} - L_{2}$ .

Substituting the values of $L_{1}$ and $L_{2}$ gives $L^{'} = \frac{g}{4 π ^{2}} (T_{1}^{2} - T_{2}^{2})$ .

The time period $T^{'}$ of the new pendulum is $T^{'} = 2 π \frac{L ^{'}}{g}$ .

Substituting $L^{'}$ gives $T^{'} = 2 π \frac{g ( T _{1}^{2} - T _{2}^{2} )}{4 π ^{2} g} = T_{1}^{2} - T_{2}^{2}$ .

Answer: $T_{1}^{2} - T_{2}^{2}$

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