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Plz sove my query as soon as possible shown below

Asked by sarveshvibrantacademy | 16 Mar, 2019, 01:13: PM

Expert Answer

when a stretched string under tension is fixed at both ends, standing waves are setup.

if y is displacment at any point x, then standing wave equation is given by, y = 2a sin(kx) cos(ωt) ...........................(1)

for fundamental mode k = 2π/λ = 2π/(2L) = π/L where λ is wavelength and L is length of string.

ω = 2πf = 2π(v/λ) = πv/L , where f is frequency = v/λ and v is velocity of wave in the string.

velocity at a given point x is given by,

begin mathsize 12px style open parentheses fraction numerator begin display style d y end style over denominator begin display style d t end style end fraction close parentheses subscript m a x end subscript space equals space minus 2 omega A space sin left parenthesis k x right parenthesis space sin left parenthesis omega t right parenthesis space equals space minus 2 A πv over L space sin k x space end style

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

for a small mass of string dm at a point x, maximum kinetic energy dk is given by,

begin mathsize 12px style d k space equals space 1 half space d m space open parentheses fraction numerator d y over denominator d t end fraction close parentheses subscript m a x end subscript superscript 2 space equals space 1 half mu space A squared fraction numerator 4 straight pi squared over denominator L squared end fraction space v squared space sin squared k x space d x space equals space space fraction numerator 2 straight pi squared straight A squared over denominator L squared end fraction space m space g space sin squared k x space d x end style

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

where μ is the linear mass density (mass /unit length).

we have used the relations

begin mathsize 12px style v space equals space square root of T over mu end root end style

and T = mg , to get the simplified eqn.(3)

Maximum Kinetic energy K_maxfor full string of length L =

begin mathsize 12px style 1 over k space fraction numerator begin display style 2 straight pi squared straight A squared end style over denominator begin display style L squared end style end fraction space m g integral subscript 0 superscript straight pi space sin squared k x space d left parenthesis k x right parenthesis space space space space........................ left parenthesis 4 right parenthesis end style

( integration limits : when x = 0, kx = 0 ; when x = L, kx = kL = π )

maximum kinetic energy K_max is obtained from eqn.(4) ,

begin mathsize 12px style K subscript m a x end subscript space equals space fraction numerator straight pi squared straight A squared over denominator L end fraction space m space g end style

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for first ovetone, maximum kinetic energy is calculated in similar manner by using k = (2π) / λ = (2π) / L

Answered by Thiyagarajan K | 17 Mar, 2019, 11:55: AM

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