The period of oscillation of simple pendulum increases by 20% when its length is increased by 44cm. Find its (i) initial length (ii) initial period of oscillation. (Ans: L₁ = 1m, T₁ = 2.007s)
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Let its initial length is L and initial time period of oscillation is T.
Then, T = 2π√(l/g) .......(1)
a/c to question,
The period of oscillation of simple pendulum increases by 20% when its length is increased by 44cm.
e.g., T + 20% of T = 2π√{(l + 0.44)/g}
6T/5 = 2π√{(l+0.44)/g}.....(2)
From equations (1) and (2),
5/6 = √{l/(l+0.44)}
25/36 = l/(l + 0.44)
25l + 25 × 0.44 = 36l
11 = 11l
l = 1m
Hence, initial length of pendulum is 1m
Now, T = 2π√(l/g)
Put l = 1m , g = 9.8 m/s² and π = 3.14
Then, you will get, T = 2.007 sec.
Then, T = 2π√(l/g) .......(1)
a/c to question,
The period of oscillation of simple pendulum increases by 20% when its length is increased by 44cm.
e.g., T + 20% of T = 2π√{(l + 0.44)/g}
6T/5 = 2π√{(l+0.44)/g}.....(2)
From equations (1) and (2),
5/6 = √{l/(l+0.44)}
25/36 = l/(l + 0.44)
25l + 25 × 0.44 = 36l
11 = 11l
l = 1m
Hence, initial length of pendulum is 1m
Now, T = 2π√(l/g)
Put l = 1m , g = 9.8 m/s² and π = 3.14
Then, you will get, T = 2.007 sec.
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Answer:
The period of oscillation of simple pendulum increases by 20% when its length is increased by 44cm. Find its (i) initial length (ii) initial period of oscillation. (Ans: L₁ = 1m, T₁ = 2.007s)
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