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Asked by sarveshvibrantacademy | 27 Mar, 2019, 10:56: AM
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Figure shows a string which is tied at the ceiling. Let us measure the length y  from bottom to top.
 
Also its linear mas density μ at y = 0 (bottom of the string ) is 0  and at y = L is λ and it is linearly varying.
 
At a distance y from bottom, linear mass density μ of string  =    (λ/L)y kg/m  ................................(1)
 
Tension T  at y due to the weight of the string length from 0 to y is  [ λ/(2L) ]y×y×3g  = λy2/(2L)  kg   .............................(2)
 
In the above eqn.(2) average linear mass density is multiplied to get mass. 
Mass is multiplied by 3g to get Tension because when lift accelerated upwards with acceleration 2g,
suspended mass will get acceleration 3g
 
velocity v at a distance y from bottom is given by,   begin mathsize 12px style v space equals space square root of T over mu end root space equals space square root of fraction numerator open parentheses bevelled fraction numerator lambda y squared over denominator 2 L end fraction close parentheses 3 g over denominator begin display style bevelled fraction numerator lambda y over denominator L end fraction end style end fraction end root space equals square root of fraction numerator 3 g over denominator 2 end fraction y end root end style .................................(3)
acceleration = begin mathsize 12px style fraction numerator d v over denominator d t end fraction space equals space fraction numerator d v over denominator d y end fraction cross times fraction numerator d y over denominator d t end fraction space equals space fraction numerator d v over denominator d y end fraction cross times v end style  .............................(4)
By substituting for v and derivative of v with respect to y using eqn.(3),  acceleration = begin mathsize 12px style square root of fraction numerator 3 g over denominator 2 end fraction end root space fraction numerator 1 over denominator 2 square root of y end fraction space cross times square root of fraction numerator 3 g over denominator 2 end fraction y end root space space equals space fraction numerator 3 g over denominator 4 end fraction end style  ..........................(4)
Hence acceleration is same everywhere
 
begin mathsize 12px style v space equals space fraction numerator d y over denominator d t end fraction space equals space square root of fraction numerator 3 g over denominator 2 end fraction y end root space space space space space space space space space o r space space space space space fraction numerator d y over denominator square root of y end fraction space equals space square root of fraction numerator 3 space g over denominator 2 end fraction end root space space space d t end style  ..............................( 5 )
we get time T taken by wave to travel from bottom to top, by integrating both sides of eqn.( 5 )
 
 
begin mathsize 12px style integral subscript 0 superscript L space fraction numerator d y over denominator square root of y end fraction space equals space square root of fraction numerator 3 space g over denominator 2 end fraction end root space space integral subscript 0 superscript T space d t space open square brackets 2 square root of y close square brackets subscript 0 superscript L space space space equals space square root of fraction numerator 3 space g over denominator 2 end fraction end root space space T space space space space space space space o r space space space space space T space space equals space square root of fraction numerator 8 L over denominator 3 g end fraction end root end style
Answered by Thiyagarajan K | 28 Mar, 2019, 01:31: PM

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