Question

anyone answer this question ​

Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

Two open organ pipe of lenght ${\sf L_1}$ and ${\sf L_2}$ . ${\sf (L_2>L_1)}$ produces x beats/sec . Then , speed of sound in organ pipe is ⎯

$\huge\underline{\underline{\bf \orange{Solution-}}}$

Frequency of organ pipe is given as

❏$\large{\boxed{\bf \blue{f=\dfrac{v}{2l}}}}$

f = Frequency

v = speed of sound in organ pipe

l = Lenght of pipe

$\implies{\sf f_1-f_2=x\:\:\:(Given) }$

$\implies{\sf f_1-f_2=\dfrac{v}{2L_1}-\dfrac{v}{2L_2} }$

$\implies{\sf x=\dfrac{v}{2}×\left(\dfrac{1}{L_1}-\dfrac{1}{L_2}\right) }$

$\implies{\sf x=\dfrac{v}{2}×\left(\dfrac{L_2-L_1}{L_1L_2}\right)}$

$\implies{\bf \red{v=2x\left(\dfrac{L_2-L_1}{L_1L_2}\right)} }$

$\huge\underline{\underline{\bf \orange{Answer-}}}$

Option (1)

${\bf \red{v=2x\left(\dfrac{L_2-L_1}{L_1L_2}\right)} }$ is the speed of sound in organ pipe.