Answer:
(a). The lowest frequency is 64.7 Hz.
(b). The next two resonant frequencies for the bugle are 129.4 Hz and 194.2 hz.
Explanation:
Given that,
Length = 2.65 m
Speed of sound = 343 m
We need to calculate the wavelength
Using formula of wavelength
[tex]\lambda=2l[/tex]
Put the value into the formula
[tex]\lambda=2\times2.65[/tex]
[tex]\lambda=5.3\ m[/tex]
(a). We need to calculate the lowest frequency
Using formula of frequency
[tex]f_{1}=\dfrac{v}{\lambda_{1}}[/tex]
Put the value into the formula
[tex]f_{1}=\dfrac{343}{5.3}[/tex]
[tex]f_{1}=64.7\ Hz[/tex]
(b). We need to calculate the next two resonant frequencies for the bugle
Using formula of resonant frequency
[tex]f_{2}=\dfrac{v}{\lambda_{2}}[/tex]
[tex]f_{2}=\dfrac{v}{l}[/tex]
Put the value into the formula
[tex]f_{2}=\dfrac{343}{2.65}[/tex]
[tex]f_{2}=129.4\ Hz[/tex]
For third frequency,
[tex]f_{3}=\dfrac{v}{\lambda_{3}}[/tex]
[tex]f_{3}=\dfrac{3v}{2l}[/tex]
Put the value into the formula
[tex]f_{3}=\dfrac{3\times343}{2\times2.65}[/tex]
[tex]f_{3}=194.2\ Hz[/tex]
Hence, (a). The lowest frequency is 64.7 Hz.
(b). The next two resonant frequencies for the bugle are 129.4 Hz and 194.2 hz.
The next two resonant frequencies for the bugle are 64.72Hz and 323.6Hz
The lowest frequency is expressed as:
f = v/2l
[tex]f_0=\frac{343}{2(2.65)} f_0=\frac{343}{5.3}\\f_0= 64.72 Hz[/tex]
Since the bugle is an open pipe the next two resonant frequencies are;
[tex]f_2 =3f_0=3(64.72) = 194.15Hz\\\\f_3 = 5f_0 = 5(64.72) =323.6Hz \\[/tex]
Hence the next two resonant frequencies for the bugle are 64.72Hz and 323.6Hz
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