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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.

Respuesta :

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 = [tex]\frac{v}{2L}[/tex]   ..............1

put here value in equation 1

lowest frequency will be = [tex]\frac{343}{2(0.32)}[/tex]

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

so

wavelength =  [tex]\frac{v}{f}[/tex]   ..............2

put here value

wavelength =  [tex]\frac{343}{4000}[/tex]  

wavelength = 0.08575 m

so distance =  [tex]\frac{wavelength}{2}[/tex]   ..............3

distance =  [tex]\frac{0.08575}{2}[/tex]  

distance = 0.0425 m

so distance  between adjacent anti nodes is 4.25 cm

A) The frequency of the lowest note a piccolo can play is; f = 535.94 Hz

B) The distance between adjacent antinodes for this mode of vibration is; D = 42.875 cm

Resonance calculations

A) The formula for the frequency of the lowest note a piccolo can play is gotten from the formula;

f = v/2L

Where;

  • v is speed of sound in air = 343 m/s
  • L is length given as 32 cm = 0.32 m

Thus;

f = 343/(2 * 0.32)

f = 535.94 Hz

B) Formula for wavelength is;

λ = v/f

Thus;

λ = 343/4000

λ = 0.08575 m

Now, the distance between adjacent antinodes for the given mode of vibration is;

D = λ/2

D = 0.08575/2

D = 0.042875 m

This gives; D = 42.875 cm

Read more about Resonance at; https://brainly.com/question/9898147

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