Question

A set of 56 tuning forks are so arranged in series that each fork gives 4 beats per second with the previous one. The frequency of the last fork Is 3 times that of the first. The frequency of the first fork is

• A. $110$
• B. $60$
• C. $56$
• D. $52$

Answer 1

The correct option is A $110$

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 $56Hz$ beats$3\times n$

(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 frequencies$4Hz$

Now, the difference between $56Hz4$ and $1st$ tunning frequencies$=3n-n=2n$

As per equation,

$3n-n=2n2n=55\times 4n=\frac{55\times 4}{2}=110$