when a tuning fork of unknown frequency is sounded with another tuning fork whose frequency is 384 vibrations per second ( that is , 384 hertz) , 6 beats per second are produced. when wax is attatched to the prong of the first fork, then on sounding it with it with the second, 4 beats per second are produced . determine the unknown frequency.
(plz. clearly explain. i m unable to understand the topic beats !!!)
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listen unknown fork is first fork according Q's and another for frequency is 384 and beat means difference b/w two frequency per sec 6 beat/sec means unknown fork has two option of frequency is 384+6=390 and 384-6=378 then wax loded on unknown fork means this fork freq. decreases Becaz of wax loading and acc. to Q's beats decrease to 4 beat per sec means agar beat decrease hua agar unknown freq. 378 hota frq. decrease hone par beat increase hota Becaz difference badta but 390 hz is right ans plzz mark this brainliest ans
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