When the length of a simple pendulum is decreased by 20 cm, the period changes by 10%. Find the original length of the pendulum.

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#### Solution

L_{2} = L_{1} - 0.20

T_{2} = T_{1} - 10%T_{1}

`T=2pisqrt(l/g)`

`:.TpropsqrtL`

`:.T_1/T_2=sqrt(L_1/L_2)`

`T_1/(0.9T_1)=sqrt(L_1/(L_1-0.2))`

Squaring both side

`1/0.81=L_1/(L_1-0.2)`

L_{1}-0.2=0.81L_{1}

∴L_{1}-0.81L_{1}=0.2

0.19L_{1} = 0.2

`L_1=0.20/0.19`

L_{1}=1.053m

Concept: Some Systems Executing Simple Harmonic Motion

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