Question

Ten tuning forks are arranged in increasing order of frequency in such a way that any two consecutive tuning forks produce $4$ beats per second. The highest frequency is twice that of the lowest frequency. Possible highest and lowest frequencies are

• A. $80\text{Hz}$ and $40\text{Hz}$
• B. $100\text{Hz}$ and $50\text{Hz}$
• C. $44\text{Hz}$ and $22\text{Hz}$
• D. $72\text{Hz}$ and $36\text{Hz}$

Answer 1

The correct option is D $72\text{Hz}$ and $36\text{Hz}$

We know beat frequency,
$f_b=|f_1-f_2|$
Since, $f_b=4\text{Hz}$ and tuning forks are arranged in increasing order of frequencies.
Let the lowest frequency be $f_1$ and highest frequency be $f_{10}$.
Then,
$f_2-f_1=4\text{Hz}$
$f_3-f_2=4\text{Hz}$
$f_4-f_3=4\text{Hz}$
.
.
.
$f_{10}-f_9=4\text{Hz}$
On adding all the above equations, we get,
$f_{10}-f_1=36\text{Hz}$
$\Rightarrow 2f_1-f_1=36\text{Hz}$ [given]
$\Rightarrow f_1=36\text{Hz}$
So, $f_{10}=2f_1=72\text{Hz}$

OR

Let the lowest frequency be $f_1$ and highest frequency be $f_{10}$.
Here we can consider $f_1,f_2,f_3.....f_{10}$ are in AP.
Difference between consecutive terms is $4\text{Hz}$
Given that $f_{10}=2f_1$
$n^{th}$ term in AP is $S_n=a+(n-1)d$
Here $a=f_1,n=10,d=4,S_n=f_{10}=2f_1$

$\Rightarrow 2f_1=f_1+9\left(4\right)\Rightarrow f_1=36\text{Hz}\&f_{10}=72\text{Hz}$