two open pipes A and B are sounded together such that beats are heard between the first overtone of A and second overtone of B. If the fundamental frequency of A and B is $256\,Hz$ and $170\,Hz$ respectively, then the beat frequency heard is
A. $4\,Hz$
B. $3\,Hz$
C. $2\,Hz$
D. $1\,Hz$

Hint: In order to solve this question we need to understand sound waves. Sound waves are longitudinal waves which require a material medium to travel, it can be considered as consecutive phenomena of compression and rarefaction. Compression means when the atoms while vibrating bring very close to each other and rarefaction involves atoms at max separation during vibration. So when two sound waves travel through medium then due to different patterns of vibration (identified through frequency) beats could be heard. So beats are different between frequencies of sound waves.

Complete step by step answer:
According to formula of overtone, an “n” overtone frequency is defined as,
${f_n} = (n + 1){f_0}$
Where ${f_n}$ ”n” overtone frequency and ${f_0}$ is fundamental frequency
Using this formula, First overtone frequency of A is, ${f_1} = 2{f_{0A}}$
Here, ${f_{0A}} = 256Hz$
Putting formula we get, ${f_1} = 2 \times 256Hz$
${f_1} = 512Hz$
Similarly, Second overtone frequency of B is, ${f_2} = 3{f_{0B}}$.

Here, ${f_{0B}} = 170\,Hz$
Putting values we get,
${f_2} = 3 \times 170\,Hz$
$\Rightarrow {f_2} = 510\,Hz$
So beat frequency is defined as difference in frequency of sound waves
Let the beat frequency be $f$
So, According to formula we get, $f = {f_1} - {f_2}$
Putting values we get, $f = (512 - 510)\,Hz$
$\therefore f = 2\,Hz$

So the correct option is C.

Note:It should be remembered that tone of sound wave is defined as sound that can be recognized by its regularity of vibration, so a simple tone is fundamental frequency while if the sound waves vibrate with more frequency then it is referred as overtone. Also an open organ pipe is defined as a pipe having both ends open and standing sound waves generate it in having node and antinode.