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Asked by sarveshvibrantacademy | 06 Apr, 2019, 03:00: PM

Expert Answer

Let y be the distance measured from top of the rope towards bottom as shown in figure.

Tension T(y) at a point y of a rope is given by, T(y) = Mg(L-y)/L .............................. (1)

where M is mass of rope and L is length of rope.

Veleocity of transverse wave, v(y) =

begin mathsize 12px style square root of fraction numerator T left parenthesis y right parenthesis over denominator mu end fraction end root space equals space square root of fraction numerator M space g space open parentheses L minus y close parentheses over denominator L cross times begin display style M over L end style end fraction end root space equals space square root of g open parentheses L minus y close parentheses end root end style

.........................(2)

where μ is linear mass density i.e. mass per unit length.

let at time τ, the freely falling object meets the pulse at location y = l

then we have for freely falling object, (1/2)gτ²= l or

begin mathsize 12px style tau space equals space square root of fraction numerator 2 space l over denominator g end fraction end root end style

......................(3)

Time τ required for the pulse to reach the location y = l is obtained from eqn.(2) as follows

begin mathsize 12px style v left parenthesis y right parenthesis space equals space fraction numerator d y over denominator d t end fraction space equals space square root of g open parentheses L minus y close parentheses end root space space space space space o r space space space space d t space equals space fraction numerator 1 over denominator square root of g end fraction space space fraction numerator d y over denominator square root of L minus y end root end fraction end style

................................(4)

integrating both sides of eqn.(4), we get,

begin mathsize 12px style tau space equals space integral subscript 0 superscript tau d t space equals fraction numerator 1 over denominator square root of g end fraction space integral subscript 0 superscript l fraction numerator d y over denominator square root of L minus y end root end fraction space equals space fraction numerator 2 over denominator square root of g end fraction open square brackets square root of L space minus space square root of L minus l end root space close square brackets end style

.......................(5)

By equating eqn.(3) and eqn.(5), and solving for l in terms of full length of rope L, we get l = (8/9)L

( user can verify the algebra, it is so long and difficult to enter all the steps involved )

hence from bottom, the pulse and the freely falling object meets at a distance (1/9)L

Answered by Thiyagarajan K | 07 Apr, 2019, 10:23: AM

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