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56 tuning forks are arranged in a series such that each fork gives 6 beats per sec with the previous one. Assuming the frequency of the last fork to be double of the first fork, the frequency of the last fork should be
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56 tuning forks are arranged in a series such that each fork gives 6 b...
Turning fork = 56turning
Beats number=4beats/s
A frequency of the last fork
Number of beats in each second= difference between frequency=4Hz
The frequency of the tuning fork of 56Hz beats3×n
(When n be the frequency of the first beat of the tuning fork)
Now, There are 55 difference between a set of 56 tunning fork.
The difference between two frequencies4Hz
Now, the difference between 56Hz4 and 1st tunning frequencies=3n−n
= 2n
As per equation,
3n−n=2n
2n=55×4
n = 55×4/2
= 110Hz
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56 tuning forks are arranged in a series such that each fork gives 6 beats per sec with the previous one. Assuming the frequency of the last fork to be double of the first fork, the frequency of the last fork should be for Class 11 2026 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 56 tuning forks are arranged in a series such that each fork gives 6 beats per sec with the previous one. Assuming the frequency of the last fork to be double of the first fork, the frequency of the last fork should be covers all topics & solutions for Class 11 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 56 tuning forks are arranged in a series such that each fork gives 6 beats per sec with the previous one. Assuming the frequency of the last fork to be double of the first fork, the frequency of the last fork should be.
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