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Waves Part 1

Section: Physics

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Question : 80 of 100

Marks: +1, -0

A cylindrical tube, open at both ends, has a fundamental frequency,

f

, in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the aircolumn is now:

$f$
$\frac{f}{2}$
$\frac{3}{4}$
$2f$

Solution:

The fundamental frequency of open tube

v_0 = \; \frac{v}{2 l_0} \;\; \dots (i)

That of closed pipe

v_c = \; \frac{v}{4 l_c} \;\; \dots (ii)

According to the problem

l_c = \; \frac{l_0}{2}

Thus

v_c = \; \frac{v}{\frac{l_0}{2}} \Rightarrow v_c \; \frac{v}{2 l} \;\; \dots

(iii)

From equations

(i)

and

(iii)

v_0 = v_c

Thus,

v_c = f \;\; (.

as

v_0 = f

is given

)

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