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Asked by miaowmira1900 | 10 Sep, 2019, 04:18: AM
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Amplitude Ar of reflected wave :-   begin mathsize 12px style A subscript r space equals space fraction numerator v subscript 2 space minus space v subscript 1 over denominator v subscript 1 plus v subscript 2 end fraction A subscript i space equals space fraction numerator 10 minus 15 over denominator 10 plus 15 end fraction cross times 5 space equals space minus 1 space m m space end style
where Ai is amplitude of incident wave, v1 is velocity in string-1 (incident wave string ) and v2 is velocity in string-2
 
Amplitude At of transmitted wave :-   begin mathsize 12px style A subscript t space equals space fraction numerator 2 v subscript 2 over denominator v subscript 1 plus v subscript 2 end fraction A subscript i space equals space fraction numerator 2 cross times 10 over denominator 10 plus 15 end fraction cross times 5 space equals space 4 space m m space end style
frequency in both the strings are same.
 
wavelength in string-1 is 1 cm . let wave length in string is x cm
 
then we have,   ( v1 / 1 )  = ( v2 / x )     or  x = ( v2 / v1 ) = 10/15 ≈ 0.67 cm
 
Hence size of transmitted pulse  (h, w) :-  ( 4 mm,  0.67 cm )
 
Hence size of reflected pulse  (h, w) :-  ( -1 mm,  1 cm )
Answered by Thiyagarajan K | 10 Sep, 2019, 14:34: PM

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