1.

A tube open at both ends has length 47 cm. Calculate the fundamental frequency of air column.(Neglect end correction. Speed of sound in air is 3.3 x102 m/s).

Answer»

Given : L = 47 cm = 47 x 10-2 m, v = 3.3 x 102 m/s

To find : Fundamental frequency of air column (n)

Formula: n = \(\frac{v}{2L}\)

Calculation: From formula,

n = \(\frac{3.3\times10^2}{2\times47\times10^{-2}}\)

\(\frac{330}{94}\times10^2\)

= antilog [log(330) – log(94)] x 102

= antilog[2.5185 – 1.9731] x 102 

= antilog[0.5454] x 102 

= 3.512 x 102

∴ n = 351.2 Hz

Ans: The fundamental frequency of air column is 351.2 Hz.



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