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  • A closed organ pipe has length ‘l'. The air in it is vibrating in 3rd overtone with maximum amplitude 'a'. Find the amplitude at a distance of l /7 from the closed end o

A closed organ pipe has length ‘l'. The air in it is vibrating in 3rd overtone with maximum amplitude 'a'. Find the amplitude at a distance of l /7 from the closed end of the pipe.       

Option: 1

a


Option: 2

\frac{a}{2}


Option: 3

\sqrt{3}\frac{a}{2}   


Option: 4

\frac{a}{\sqrt{2}}


Answers (1)

best_answer

 

As we have learnt in

 

The frequency in closed organ pipe -

\nu = \left ( 2n+1 \right )\frac{V}{4l}

n= 0,1,2,3...............
 

- wherein

V= velocity of sound wave

l= length of pipe

n= number of overtones

 

 The figure shows variation of displacement of particles in a closed organ pipe for 3rd overtone.

For third overtone l=\frac{7\lambda }{4}\; or\; \lambda =\frac{4l}{7}\; or\; \frac{\lambda }{4}=\frac{l}{7}

               

Hence the amplitude at P at a distance  \frac{l}{7} from closed end is ‘a’ because there is an antinode at that point

Alternate

            Because there is node at x = 0 the displacement amplitude as function of x can be written as

A=a \; sin\; kx=a\; sin\; \frac{2\pi }{\lambda }X

For  third overtone  l=\frac{7\lambda }{4}\; or\; \lambda =\frac{4l}{7}

\therefore A=a\; sin\; \frac{7\pi }{2l}\; \frac{l}{7}=a\;\; sin\; \frac{\pi }{2}=a                          at x=\frac{l}{7}\Rightarrow A=a

Posted by

Rakesh

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