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Asked by sarveshvibrantacademy | 16 Mar, 2019, 01:13: PM
answered-by-expert 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 Kmax for 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 Kmax 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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