56 tuning forks are so arranged in increasing order of frequencies in series that each fork gives 4

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14.Waves and Sound

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$56$ tuning forks are so arranged in increasing order of frequencies in series that each fork gives $4$ beats per second with the previous one. The frequency of the last fork is the octave of the first. The frequency of the first fork is ..... $Hz$

A

$220$

B

$224$

C

$(220/7)$

D

$110$

Solution

Frequency of first fork $=n .$ Frequency of 56 th fork $=2 n$

$\therefore n+(56-1) 4=2 n$

$\therefore n+220=2 n$

$n=220 H z$

Standard 11

Physics

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