# Frequency of sound in a pipe question

## Homework Statement

Shown in the picture attachment.

## The Attempt at a Solution

Hello! with this question I am just completely lost. I haven't the slightest clue how to get started. Anybody care to explain why the answer is D? (I've looked at the markscheme..)

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## Homework Statement

Shown in the picture attachment.

## The Attempt at a Solution

Hello! with this question I am just completely lost. I haven't the slightest clue how to get started.

their will be loud sound when stationary waves will be formed...(it's key point for this question)
for such cases(when one end closed and one end open)
f=(4n+1)c/(4L)
for both the cases f is constant c(speed of sound) is constant.
in first cases take n=k and in second case take it k+1. you will have two equations having a single variable k. just eliminate it and you will get speed of sound in terms of L1 L2 and f.
Anybody care to explain why the answer is D? (I've looked at the markscheme..)
it's bad habit. looking at answer and trying to somehow explain that...
I have not done it. but i think my solution will work.

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Hi! being quite a simpleton, I found the answer a bit confusing. I understand that as the pipe is lowered standing waves will be produced... I also get that f is constant as the tuning fork held determines what possible lengths that cause the louder sounds. Why will the frequency be equal to (4n+1)c/4L then though? I'd figure that the wavelength at L1 of the sound in the pipe would be 4L, but then how does that translate into the equation you presented? Sorry i'f i'm a little inept! Thankyou for your help thus far.

NOTE: i have done an error in my last post formula is not f=(4n+1)c/(4L) it is f=(2n+1)c/(4L)..
Why will the frequency be equal to (4n+1)c/4L then though?

You might have seen it's derivation in your book...
there will be a node at closed end and anti node and open end..
I'd figure that the wavelength at L1 of the sound in the pipe would be 4L
that is not necessary...
what if is say it is 4L/3. or 4L/5..
that can be any out of this series..

λ/4=L(fundamental wave)
3λ/4=L (first overtone)
5λ/4=L(second overtone)
..
..
..
(2n+1)λ/4=L (replace λ by c/f to get frequency formula as i have written in my last post)

but as it is stated in question that large sound listened first time when distance is L1 and very next is L2 so if in case of L1 n=k then in case of L2 it is (k+1)..
if you are still not clear that what's happening then once again read this topic from your book ...

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Resonance occurs for pipe lengths such that there is a displacement node at the closed end and a displacement antinode at,approximately,the open end.First resonance occurs for a length of L1 when 1/4 of the wave fits in the space.Second resonance occurs for a length of L2 when 3/4 of a wave fits in the space.From that you can work out the wavelength in terms of L1 and L2 and hence the speed.
To see it clearly sketch it out.

Aha! I follow. so it doesn't necessarily have to be the first harmonic, but they will definitely be in series? I've looked at it a little more and I think it's finally embedded itself into my brain. Thankyou for your help!