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. (b) opening holes in the side effectively shortens the length of the resonant column. if the highest note a piccolo can sound is 4,000 hz, find the distance between adjacent antinodes for this mode of vibration.

lowest frequency = 535.93 Hz

distance  between adjacent anti nodes is 4.25 cm

Explanation:

given data

length L = 32 cm = 0.32 m

to find out

frequency and distance between adjacent anti nodes

solution

we consider here speed of sound through air at room temperature 20 degree is  approximately  v = 343 m/s

so

lowest frequency will be =   ..............1

put here value in equation 1

lowest frequency will be =

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

so

wavelength =     ..............2

put here value

wavelength =    

wavelength = 0.08575 m

so distance =     ..............3

distance =    

distance = 0.0425 m

so distance  between adjacent anti nodes is 4.25 cm

that would be because of both gravity and friction. so     c