Solve Wave Problems: Calculating Length for Higher Note on Slide Whistle

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To play a note one octave higher on a 27 cm slide whistle, the length of the whistle must be halved, resulting in a new length of 13.5 cm. An octave corresponds to doubling the frequency, which also halves the wavelength. The speed of sound is given as 343 m/s, and relevant equations relate velocity, wavelength, and frequency. Understanding the relationship between tube length and wavelength is crucial, as the length of the whistle is directly proportional to the wavelength. Thus, the correct length for the higher note is 13.5 cm.
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Homework Statement


A slide whistle has a length of 27 cm. If you want to play a note one octave higher, the whistle should be how long?

Homework Equations



velocity=(wavelength)*(frequency)
speed of sound= 343 m/s

The Attempt at a Solution



343=6.75f
50.8=f

50.8*2= 101.63Hz

( i am so lost)
 
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Q#1: What happens to the frequency when something goes one octave higher?
Q#2: What happens to the wavelength when {insert answer to Q#1 here} happens to the frequency?
 
it doubles?
 
Yes. What happens to the wavelength when the frequency doubles?
 
it halves?

so is the answer 101.63? or 27/2?
 
manda said:

Homework Statement



A slide whistle has a length of 27 cm. If you want to play a note one octave higher, the whistle should be how long?

Homework Equations



velocity=(wavelength)*(frequency)
speed of sound= 343 m/s

The Attempt at a Solution



343=6.75f
50.8=f

50.8*2= 101.63Hz

( i am so lost)


Hi Manda, I saw you cry for help. I'll give it a shot.

An octave is the interval between one musical pitch and another with half or double its frequency. For example, if one note has a frequency of 400 Hz, the note an octave above it is at 800 Hz, and the note an octave below is at 200 Hz.

Further octaves of a note occur at 2^n times the frequency of that note (where n is an integer), such as 2, 4, 8, 16, etc. and the reciprocal (1/2^n) of that series. For example, 50 Hz and 400 Hz are one and two octaves away from 100 Hz because they are ½ (or 2^−1) and 4 (or 2^2) times the frequency, respectively.

So first calculate the lowest frequency of the note in the whistle, using your relevant equations, as well as the length of the whistle. Then double that frequency to get the octave one higher. Let me know if I should clarify. ~M
 
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Yes. One of those two is correct; which is referring to a length? (Since the question is asking for a length).

p.s. It might also help to find the equation involving whistle (or tube) length. Is there a discussion on tubes in your book, in the chapter dealing with sound or waves?
 
27/2

umm yeah there is 2 paragraphs
all they say is open = Y/2
closed=y/4 and then y/2


does that sound right?
 
Yes, that's right. In all those cases, the tube length is proportional to the wavelength.
So when the wavelength halves, the tube length _____ ?
 
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