Answer:
68.8 Hz
137.6 Hz, 206.4 Hz
Explanation:
L = Length of tube = 2.5 m
v = Velocity of sound in air = 344 m/s
Distance between nodes is given by
[tex]L=\dfrac{\lambda}{2}+m\dfrac{\lambda}{2}\\\Rightarrow \dfrac{\lambda(n+1)}{2}=L\\\Rightarrow \lambda=\dfrac{2L}{n+1}[/tex]
Where n = 0, 1, 2, 3, ...
Making n+1 = n
[tex]\lambda=\dfrac{2L}{n}[/tex]
where n = 1, 2, 3 .....
For fundamental frequency n = 1
[tex]\lambda=\dfrac{2\times 2.5}{1}\\\Rightarrow \lambda=5\ m[/tex]
Frequency is given by
[tex]f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{344}{5}\\\Rightarrow f=68.8\ Hz[/tex]
The fundamental frequency is 68.8 Hz
First overtone
[tex]2f=2\times 68.8=137.6\ Hz[/tex]
Second overtone
[tex]3f=3\times 68.8=206.4\ Hz[/tex]
The overtones are 137.6 Hz, 206.4 Hz