The overall length of a piccolo is 30.0 cm. The resonating air column is open at both ends. (a) Find the frequency of the lowest note a piccolo can sound. (Assume that the speed of sound in air is 343 m/s.) Hz (b) Opening holes in the side of a piccolo effectively shortens the length of the resonant column. Assume the highest note a piccolo can sound is 3 000 Hz. Find the distance between adjacent antinodes for this mode of vibration.

(a)  the frequency of the lowest note the piccolo can sound is 571.7 Hz

(b) the distance between adjacent antinodes is 5.72 cm

Explanation:

(a)

Given;

length of piccolo, L = 30 cm = 0.3 m

the speed of sound in air is 343 m/s

The wavelength of a pipe open at both ends, for the first harmonic is given;

L = A → N + N → A

L = λ / 4 +  λ / 4

L = λ / 2

λ = 2L

λ = 2 x 0.3 = 0.6 m

The fundamental frequency (lowest frequency) is given  by;

f₀ = v / λ

f₀ = (343 / 0.6)

f₀ = 571.7 Hz

(b)

Given;

highest note, f = 3000 Hz

the distance between adjacent antinodes is given by;

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

the assestment will probley apele with cause and effect