Question

discuss the formation of standing waves in closed pipe. hence show that the ratio of frequency is 1:3:5 ​

Answer 1

Answer:

To prove:

Formation of standing waves in a closed pipe is in the ratio of 1:3:5 .....

Concept:

Closed pipe refers to a pipe which is closed on 1 side and open on the other side. This means that on the open side , there will be formation of anti-nodes and the closed sides will form nodes.

Calculation:

In the 1st case :

$\therefore \: \frac{ \lambda}{4} = l \\ = > \lambda = 4l$

$freq_{1} = \dfrac{velocity}{wavelength} \\ = > freq_{1} = \dfrac{v}{4l} ...............(1)$

For 2nd case :

$\therefore \: \frac{ \lambda}{4} + \frac{ \lambda}{2} = l \\ = > \lambda = \frac{4l}{3}$

$freq_{2}= \dfrac{velocity}{wavelength} \\ = > freq_{2} = \dfrac{3v}{4l} ...............(2)$

For 3rd case :

$\therefore \: \frac{ \lambda}{4} + \frac{ \lambda}{2} + \frac{ \lambda}{2} = l \\ = > \lambda = \frac{4l}{5}$

$freq_{3}= \dfrac{velocity}{wavelength} \\ = > freq_{3} = \dfrac{5v}{4l} ...............(3)$

So the required ratio are :

$\boxed{ \huge{ \red{1 \: : 3 \: : 5}}}$

Answer 2

$\huge {\red{\boxed{ \overline{ \underline{ \mid\mathfrak{An}{\mathrm{sw}{ \sf{er}} \colon\mid}}}}}}$

Formula for Wavelength of Closed organ pipe is :

$\large{\boxed{\boxed{\sf{\lambda \: = \: \dfrac{4L}{2n \: - \: 1}}}}}$

First Mode of Vibration

Put n = 1

$\implies {\sf{λ_1\: = \: \dfrac{4L}{2(1) - 1}}} \\ \\ \implies {\sf{λ_1 \: = \: \dfrac{4L}{2-1}}} \\ \\ \implies {\sf{λ_1 \: = \: \dfrac{4L}{1}}} \\ \\ \implies {\sf{L \: = \: \dfrac{λ_1}{4}}} \\ \\ {\boxed {\sf{\nu _1 \: = \: \dfrac{v}{4L}}}}$

__________________________________

Second Mode of Vibration

Put n = 2

$\implies {\sf{λ_2 \: = \: \dfrac{4L}{2(2) \: - \: 1}}} \\ \\ \implies {\sf{λ_2 \: = \: \dfrac{4L}{4 \: - \: 1}}} \\ \\ \implies {\sf{λ_2 \: = \: \dfrac{4L}{3}}} \\ \\ {\boxed{\sf{\nu _2 \: = \: 3 \nu _1}}}$

___________________________

Put n = 3

$\implies {\sf{λ_3 \: = \: \dfrac{4L}{2(3) \: - \: 1}}} \\ \\ \implies {\sf{λ_3 \: = \: \dfrac{4L}{6 \: - \: 1}}} \\ \\ \implies {\sf{λ_3 \: = \: \dfrac{4L}{5}}} \\ \\ {\boxed{\sf{\nu _3 \: = \: 5 \nu _1}}}$

__________________________________

For Ratio Divide all the Frequencies :

$\implies {\sf{Ratio \: = \: \dfrac{\nu_1}{\dfrac{3 \nu_1}{5 \nu _1}}}} \\ \\ \implies {\sf{Ratio \: = \: \dfrac{1}{\dfrac{3}{5}}}} \\ \\ \implies {\sf{Ratio \: = \: 1:3:5}}$