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for resonance in closed column freq. = (2p+1)*vel/4l . but in class 10 i learned for resonance with a tuning fork say 512 hz this value of frequency should be 512 hz only . so my question if i put p =1,2,3... how to prove that the frequency is remaining same .
 one year ago
 one year ago
for resonance in closed column freq. = (2p+1)*vel/4l . but in class 10 i learned for resonance with a tuning fork say 512 hz this value of frequency should be 512 hz only . so my question if i put p =1,2,3... how to prove that the frequency is remaining same .
 one year ago
 one year ago

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mailtoarkoBest ResponseYou've already chosen the best response.1
some one please comment what ever you feel about my question
 one year ago

mailtoarkoBest ResponseYou've already chosen the best response.1
i know the length is changing but how to mathematically get it proved that the freq. is remaining constant
 one year ago

mailtoarkoBest ResponseYou've already chosen the best response.1
come on frnds anybody reply
 one year ago
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