The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. a. Find the frequency of the lowest note a piccolo can play, assuming the speed of sound in air is 340 m/s.

 

b. Opening holes in the side effectively shortens the length of the resonant column. If the highest note a piccolo can sound is 4000 Hz, find the distance between adjacent antinodes for this mode of vibration.

 

Open-Open system, so Lambda=2L (1st Harmonic, Am I right here?) =2(0.32m)=0.64m

Frequency=340m/s / 0.64m=531.25Hz ----------->a. Is that right?

 

b) F=4000Hz, Distance=x, antinode = 1/2 of Lambda? so 0.16m ------------> Is that right?

I am confused at this point.

The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. a. Find the frequency of the lowest note a piccolo can play, assuming the speed of sound in air is 340 m/s.

 

b. Opening holes in the side effectively shortens the length of the resonant column. If the highest note a piccolo can sound is 4000 Hz, find the distance between adjacent antinodes for this mode of vibration.

 

Open-Open system, so Lambda=2L (1st Harmonic, Am I right here?) =2(0.32m)=0.64m

Frequency=340m/s / 0.64m=531.25Hz ----------->a. Is that right?

Can't see anything wrong with your logic. And that matches quite closely to the tuning of a real piccolo (roughly 525-550Hz depending on whether it's C or D flat).

b) F=4000Hz, Distance=x, antinode = 1/2 of Lambda? so 0.16m ------------> Is that right?

I am confused at this point.

Well step 1 would be to find the wavelength that corresponds to this new frequency. There will be more than one full wave in the flute. It will (barring complicated effects we're ignoring), at least, have a pressure node at the open hole closest to the mouth-piece. So we expect it to be a lot shorter.

 

[math]v=f\lambda \rightarrow 340ms^{-1}=4000s^{-1}\lambda[/math]

Then the distance between two nodes/antinodes would be half of that.