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NEET Class neet Answered

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Asked by brijk456 | 02 Oct, 2019, 11:11: PM
answered-by-expert Expert Answer
Frequency of first tuning fork = 256 Hz.
 
Frequency of last tuning fork = 512 Hz.
 
Successive tuning fork produce 4 beats, hence frequency difference between successive tuning forks = 4 Hz.
 
frequncies of all tuning forks are in Arithmetic progression,
 
first value is 256 Hz, common difference 4 Hz and the last one is 512 Hz.
 
Hence the number of tuning forks n is obtained form :-    256 + ( n - 1) × 4 = 512  ...............(1)
 
Hence from eqn.(1), we get n = 65
Answered by Thiyagarajan K | 03 Oct, 2019, 12:07: AM
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