Explanation:
The frequency of an organ pipe if it is open is given by :
[tex]f=\dfrac{nv}{2l}[/tex]
v is speed of sound in air is 343 m/s at 20°C
For fundamental frequency, n = 1
[tex]f=\dfrac{1\times 343}{2\times 0.26}\\\\f=659.61\ Hz[/tex]
First overtone frequency,
[tex]f_1=2f\\\\f_1=2\times 659.61\\\\f_1=1319.22\ Hz[/tex]
Second overtone frequency,
[tex]f_2=3f\\\\f_2=3\times 659.61\\\\f_2=1978.83\ Hz[/tex]
Third overtone frequency
[tex]f_3=4f\\\\f_3=4\times 659.61\\\\f_3=2638.44\ Hz[/tex]
Hence, this is the required solution.
