two parts of the sonometer wire,divided by a movable knife-edge,differ in length by 1cm and produce one beat per second when sounded together.if the totall length of the wire is 100cm,the frequencies of the two parts of the wire a

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Solution

$Total length, L = 100 cm Difference of lengths of two parts = 1 cm Lengths of the parts = 50.5 cm and 49.5 cm As the frequency of the string is inversely proportional to length, the frequencies are 49.5 Hz and 50.5 Hz$

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two parts of the sonometer wire,divided by a movable knife-edge,differ in length by 1cm and produce one beat per second when sounded together.if the totall length of the wire is 100cm,the frequencies of the two parts of the wire a

Total length, L=100 cmDifference of lengths of two parts=1 cmLengths of the parts=50.5 cm and 49.5 cmAs the frequency of the string is inversely proportional to length,the frequencies are 49.5 Hz and 50.5 Hz

$Total length, L = 100 cm Difference of lengths of two parts = 1 cm Lengths of the parts = 50.5 cm and 49.5 cm As the frequency of the string is inversely proportional to length, the frequencies are 49.5 Hz and 50.5 Hz$