CBSE Class 12-science Answered
Fifty six tuning forks are arranged in a series such that each fork gives 6 beats per second with the preceding one. Assuming the frequency of the last fork to be double of the first fork, the frequency of the last fork should be
(a) 220Hz (b) 330Hz (c) 440Hz (d) 660Hz
Asked by sunil2791 | 25 Apr, 2018, 07:01: PM
Expert Answer
56 tuning forks arranged in series in increasing order of frequency.
Each fork gives 6 beats with the preceeding one means, frequency of each fork is 6 more than the preceeeding one.
Hence 56 frequencies are in arithmatic progression with common difference is 6 Hz.
it is given that frequency of last one f56 is double of first one f1 ;
f56 = 2×f1 = f1+55×6 ; this gives f1 = 330 Hz ; hence last one is 660 Hz
Answered by Thiyagarajan K | 26 Apr, 2018, 12:37: PM
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