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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: danielf on October 27, 2008, 08:29:25 am

Title: Harmonics
Post by: danielf on October 27, 2008, 08:29:25 am: Came across a question in the 2006 NEAP Trial Exam and wasn't quite sure if the answer given was wrong:

A tube of length 0.765m, closed at one end and with a speaker at the open end, is used in a sound experiment. The freq. is increased from 0Hz until a microphone situated 0.085m from the closed end detects a maximum intentsity of sound.
What number harmonic is present when the microphone detects maximum amplitude for the first time?

My working showed that, if an antinode is to exist at both the closed end and the point where the mic is, it must be the 18th harmonic (which I know is not possible because only odd harmonics are possible). The answer says it is the 5th harmonic (b/c they say 2.5 wavelengths are present in the tube) but no matter how I draw it there always seems to be a node at the closed end.

Help?!? :tickedoff:
Title: Re: Harmonics
Post by: cara.mel on October 27, 2008, 08:46:24 am: ignore this post it is wrong

(http://img.photobucket.com/albums/v711/happypuff/physics2.jpg)

Let me know if I havent explained myself enough
Title: Re: Harmonics
Post by: danielf on October 27, 2008, 09:05:03 am: Sorry I didn't explain the question properly - The microphone is inside the tube (ie 0.68m from the open end) - Hope that helps
Title: Re: Harmonics
Post by: cara.mel on October 27, 2008, 09:14:09 am: yeah my picture drawing ability fails too :( I can't count. I need to wake up. ignore it

I think this is right:
(http://img.photobucket.com/albums/v711/happypuff/physics3-1.jpg)
Title: Re: Harmonics
Post by: danielf on October 27, 2008, 12:32:57 pm: Really sorry to annoy and really appreciate the help but I thought that maximum intensity always occured at the closed end (the pic has a node at the closed end) which would mean there would have to be 0.5 wavelengths between the mic and the end? - But if you do this you get a max intensity at the open end as well?!?! wtf!!
Title: Re: Harmonics
Post by: cara.mel on October 27, 2008, 12:40:49 pm: yeah it does
I've just dug a big hole for myself :(
Title: Re: Harmonics
Post by: Mao on October 27, 2008, 03:14:27 pm: haha, I remember doing this question. the answer is quite simply the person who wrote it was probably drunk at the time.

it is absolutely stuffed =] so dw about it.

thought you might want my reasoning:

it is a closed pipe, and a maximum intensity is observed 0.085 from the closed end, meaning that a multiple of $\frac{\lambda}{2}$ exist within this space.

now, $\frac{0.765}{0.085}=9$ , hence we grab our multiples of $\frac{\lambda}{2}$ and copypasta it 8 times and fit it inside this closed tube.

the dilemma is, no matter how many multiples of these we choose, we will ALWAYS end up with an antinode at the open end because of how intricately it all piece together, i.e. it is impossible. [well, the reason is that a closed-at-one-end pipe characteristically has one less $\frac{\lambda}{4}$ , which is unachievable as we have a whole number multiple of $\frac{\lambda}{2}$ . like trying to get to an odd number by adding even numbers together]
Title: Re: Harmonics
Post by: danielf on October 27, 2008, 03:40:39 pm: Lol that makes perfect sense thankyou so much I was freakin out before but this clears things up
Title: Re: Harmonics
Post by: onlyfknhuman on October 27, 2008, 07:28:50 pm: Quote from: Mao on October 27, 2008, 03:14:27 pm
haha, I remember doing this question. the answer is quite simply the person who wrote it was probably drunk at the time.

it is absolutely stuffed =] so dw about it.

thought you might want my reasoning:

it is a closed pipe, and a maximum intensity is observed 0.085 from the closed end, meaning that a multiple of $\frac{\lambda}{2}$ exist within this space.

now, $\frac{0.765}{0.085}=9$ , hence we grab our multiples of $\frac{\lambda}{2}$ and copypasta it 8 times and fit it inside this closed tube.

the dilemma is, no matter how many multiples of these we choose, we will ALWAYS end up with an antinode at the open end because of how intricately it all piece together, i.e. it is impossible. [well, the reason is that a closed-at-one-end pipe characteristically has one less $\frac{\lambda}{4}$ , which is unachievable as we have a whole number multiple of $\frac{\lambda}{2}$ . like trying to get to an odd number by adding even numbers together]

how do you know a multiple of lambda/2 exists within the space of .085? i dont get it ;S
Title: Re: Harmonics
Post by: danielf on October 28, 2008, 10:27:06 pm: Does the pic below help?
There's 9 equal parts (0.765/0.085) and the 0.085m to the end is one of those parts.
To have a max at both the end and the mic you need 0.5lambda in between (by inspection of the picture)
Cos the questions so stuffed i don't think any equations will get you to knowing its 1/2 lambda
Title: Re: Harmonics
Post by: Mao on October 28, 2008, 10:52:01 pm: closed end is an antinode, and a maximum is observed 0.085 away from the closed end (i.e. another antinode)

hence, it has to be at least half a period (or multiples of such)
Title: Re: Harmonics
Post by: onlyfknhuman on November 01, 2008, 03:47:32 pm: alright i understand