Question

$41$ forks are so arranged rthat each produces $5$ beats per sec when sounded with its near fork . If the frequency of last fork is double the frequency of first fork, then the frequency if the first and last fork are respectively

• A. $200,400$
• B. $205,410$
• C. $195,390$
• D. $100,200$

Answer 1

The correct option is A $200,400$

Given 41 forks such that when two adjacent forks are struck they produces 5 beats per second

with the last fork having frequency double of the first fork.

Lets suppose that the first fork has a frequency of $f$

Then the $2_{nd}$ fork can have a frequency of $f+5orf-5$

but, the frequency of last fork is double (greater) than the frequency of first

fork which means that the frequencies are gradually increasing as we go from first to the last fork.

So, the frequency of $2_{nd}$ fork is $f_2=f+5$

the $3_{rd}$ fork is $f_3=f+10$ and so on because $f_{beat}=5$ for adjacent elements.

This sequence constitutes an arithmetic progression with $a_1=f\&d=5$

So, frequency of the 41st fork is $f_{41}=f+(41-1)5=f+200$

$f_{41}=2\times f$

$f+200=2f\Rightarrow f=200$

$f_{41}=f+200=200+200=400$

So A is the correct answer.