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Solve

Asked by adjacentcalliber10 | 14 May, 2019, 11:12: AM

Expert Answer

Initial time period of the system is,

T space equals space 2 straight pi space square root of M over k end root

... (1)

When the mass is increased by m, time period becomes 5T/3,

Thus,

fraction numerator 5 T over denominator 3 end fraction space equals space 2 pi space square root of fraction numerator M plus m over denominator k end fraction end root space... space left parenthesis 2 right parenthesis T h u s comma B y space d i v i d i n g space t h e space e q u a t i o n space left parenthesis 2 right parenthesis space b y space left parenthesis 1 right parenthesis comma w e space g e t comma fraction numerator fraction numerator 5 T over denominator 3 end fraction over denominator T end fraction space equals space fraction numerator 2 pi space square root of fraction numerator M plus m over denominator k end fraction end root over denominator 2 pi space square root of M over k end root end fraction

Solving this,

5 over 3 equals square root of fraction numerator M plus m over denominator M end fraction end root 25 over 9 equals fraction numerator M plus m over denominator M end fraction

Thus,

25 M = 9M + 9m

25 M - 9 M = 9m

Thus,

9 m = 16 M

Given m = 16 g

Thus,

16 M = 9 x 16

M space equals space fraction numerator 9 space cross times space 16 over denominator 16 end fraction

M = 9 g

Thus, value of M = 9 g

Answered by Shiwani Sawant | 14 May, 2019, 12:19: PM

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