Question

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.

Answer 1

Answer:

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 = $\frac{v}{2L}$   ..............1

put here value in equation 1

lowest frequency will be = $\frac{343}{2(0.32)}$

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

so

wavelength =  $\frac{v}{f}$   ..............2

put here value

wavelength =  $\frac{343}{4000}$  

wavelength = 0.08575 m

so distance =  $\frac{wavelength}{2}$   ..............3

distance =  $\frac{0.08575}{2}$  

distance = 0.0425 m

so distance  between adjacent anti nodes is 4.25 cm