Respuesta :
Explanation:
For a pipe with one end open ,we have the formula for fundamental frequency as
[tex]f_1= \frac{v}{4L}[/tex]
v= velocity of sound in air =340 m/s
L= length of pipe
hence, [tex]L= \frac{v}{4f_1}[/tex]
given f_1 = 440 Hz
Substituting , L = 340/(4×440) = 0.193 m
a) Let the bottle be filled to height , h.
Given that height of bottle ,H = 25 cm = 0.25 m
Also we found out that length of pipe L = 0.193 m
So, we have h = H - L
= 0.25 - 0.193 = 0.057 m = 5.7 cm.
Answer:
(a). The bottle filled at the height is 5.5 cm
(b). The frequency of the nest higher harmonic of this bottle is 880 Hz.
Explanation:
Given that,
Frequency = 440.0 Hz
Height of bottle = 25.0 cm
Suppose,
Let the bottle be filled to height h.
For a pipe with one end open,
We need to calculate the length of the pipe
Using formula of fundamental frequency
[tex]F=\dfrac{v}{4L}[/tex]
[tex]L=\dfrac{v}{4f}[/tex]
Where, L = length
v = speed of sound
Put the value into the formula
[tex]L=\dfrac{343}{4\times440}[/tex]
[tex]L=0.195\ m[/tex]
(a). We need to calculate the height
Using formula of height
[tex]h=H-l[/tex]
Where, H = height of bottle
l = length of pipe
Put the value into the formula
[tex]h=25.0\times10^{-2}-0.195[/tex]
[tex]h=0.055\ m[/tex]
[tex]h=5.5\ cm[/tex]
(b). We need to calculate the frequency of the nest higher harmonic of this bottle
Using formula of frequency
[tex]f_{next}=nf_{1}[/tex]
Where, [tex]f_{1}[/tex]=fundamental frequency
Put the value into the formula
[tex]f_{2}=2\times440[/tex]
[tex]f_{2}=880\ Hz[/tex]
Hence, (a). The bottle filled at the height is 5.5 cm
(b). The frequency of the nest higher harmonic of this bottle is 880 Hz.
