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प्रश्न
The period of oscillation of the simple pendulum increases by 20 %, when its length is increased by 44 cm. find its initial length.
उत्तर
Given:
`"T"_2 = (1 + 20/100)"T"_1`
∴ `"T"_2/"T"_1 = 120/100, "L"_2 = ("L"_1 + 0.44)`m
To find: Initial length (L1)
Formula: `"T" = 2pisqrt("L"/"g")`
Calculation:
From formula,
`"T"_1 = 2pisqrt("L"_1/"g")` ..........….(1)
`"T"_2 = 2pisqrt("L"_2/"g")` ..........….(2)
Dividing equation (1) by equation (2),
∴ `"T"_1/"T"_2 = sqrt("L"_1/"L"_2)`
∴ `100/120 = sqrt("L"_1/"L"_2)` ...........(Given)
∴ `10/12 = sqrt("L"_1/("L"_1 + 0.44))`
∴ 1.44 L1 = L1 + 0.44
∴ L1 = 1 m
The initial length of the pendulum is 1 m.
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