Answer:
(d) 368 Hz
Explanation:
The resonance frequency of one closed end pipe is given by [tex]f=\frac{nv}{4L}[/tex] where n =1,3,5,7------------
Here we are talking about second lowest resonant frequency so n=3
The speed of sound v = 343 m/sec
Length of pipe L =0.7 m
So [tex]f=\frac{nv}{4L}=\frac{3\times 343}{4\times 0.7}=367.5\ Hz[/tex]
Which is closest to 368 Hz so option (d) will be the correct answer
