NEET Class neet Answered
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Asked by brijk456 | 02 Oct, 2019, 11:11: PM
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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