Question

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Answer 1

An organ pipe of length 80 cm is opened at x=0 and closed at x=80 cm. Speed of sound in the air column is 320 m/s.If standing waves are generated in the closed organ pipe,then the correct equation of standing waves is are ( here s=longitudinal displacement , P_ex=pressure excess)(Neglect the end correction)

Answer 2

is it correct ?
\[\large \bf P_{ex}=A \sin(\frac{25 \pi}{8}x)\cos(1000\pi t)\]

Answer 3

@Michele_Laino and @samigupta8

Answer 4

U asked same doubt @mayankdevnani
N yep obviously dat is d ans

Answer 5

yep, but i think it is incorrect

Answer 6

So what r u getting ?

Answer 7

\[\large \bf P_{ex}=Acos(\frac{25 \pi}{8}x)\sin(1000 \pi t)\]
this what i get

Answer 8

Option me hi ni h naa shayd

Answer 9

yep :(

Answer 10

@Michele_Laino @IrishBoy123 @Vincent-Lyon.Fr

Answer 11

\( P(x,t)=A \sin(\dfrac{25 \pi}{8}x)\cos(1000\pi t)\)

is correct for the harmonic no 5.

You need a sine function since the pipe is open at x=0 (acoustic pressure is zero).
Using a cosine would produce a maximum amplitude for pressure at x=0, caracteristic of a closed end.

Answer 12

pipe is open at x=0,then P(x,t)=0

Answer 13

@Vincent-Lyon.Fr can we say that both the terms can also be of sine function .
In dis way ur argument wud also hold true...Is it necessary that one has to be cos n other a sine function....

Answer 14

If we consider the fundamental frequency, or fundamental wavelength, which is four times the length \(L\) of the pipe, then the excess of pressure \(\Delta p\), is given by the subsequent formula:
\[\Large \Delta p = - \rho v_S^2\frac{{\partial s}}{{\partial x}}\]
where
\(v_S\) is the sound speed in the air, \(\rho\) is the density of air, furthermore, we have:
\[\Large s\left( x \right) = A\sin \left\{ {2\pi \left( {\frac{{{v_S}}}{{4L}}t - \frac{x}{{4L}}} \right)} \right\}\]

Answer 15

but standing wave superposition for pressure variation is in the form:-
\[\large \bf \triangle P=\triangle P_0(\cos kx)(\sin \omega t)\]

Answer 16

but answer is in the form of :-
\[\large \bf \triangle P_0= \triangle P_0(\sin kx)(\cos \omega t)\]
How?

Answer 17

Exactly @Michele_laino so v have equation of pressure variation as
Po cos(wt-kx)....n when want to find stationary wave den one eq. Is dis n other wud be Po cos(wt+kx) for the wave travelling in opposite direction n we find superposition of the two then we will be having both terms in cos functions....

Answer 18

Whether you have a cos or a sine for the \(\omega t\) part is irrelevant.
You can have \(\cos (\omega t+\phi)\) with ANY value of \(\phi\) you want.
They are all solutions for your problem.

Answer 19

okay.What about `kx` ?
coskx or sinkx ?

Answer 20

I already answered that question :
- only sinkx is ok if x=0 is an open end (no acoustic pressure there, maximum velocity)
- only coskx is ok if x=0 is an closed end (no velocity there, maximum acoustic pressure)

Answer 21

yes! We have to find the resulting stationary wave by means of the superposition principle first @samigupta8

Answer 22

Thanks....@Michele_Laino u cleared my doubt....

Answer 23

:)

Answer 24

the pressure variation is the effect and the stationary waves are the cause

Answer 25

So they can be in any form sine or cos....v don't have to give a damn to DAT thing...

Answer 26

I think that the form can be sin.. or cos.. since both type of solution type obey to the oscillatory motion.
Furthermore we can get a sin.. solution starting from the cos.. form, depending on the initial conditions

Answer 27

Yeah...correct BT der is no other option DAT can stand for match vid harmonic frequencies ....so DAT also rules out d case

Answer 28

yes! More precisely we can say that sin.. or cos.. represent the same \(physical\) behaviour

Answer 29

Yep....

Answer 30

@Michele_Laino and @Vincent-Lyon.Fr is my answer correct also ?

Answer 31

Which one, please?

Answer 32

This one

Answer 33

No, the cosine is wrong since the question states that x=0 is an open end.

Answer 34

if we replace it by displacement, then is it true?

Answer 35

i think it would be,right ?

Answer 36

@Vincent-Lyon.Fr

Answer 37

Yes, it would be true for velocity or for displacement, but the amplitude A would be a velocity or a displacement respectively.

Answer 38

i got it now ! thank you