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BITSAT 2019 Paper

Section: Physics

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Question : 29 of 150

Marks: +1, -0

The fundamental frequency of an open organ pipe is 300 Hz. The first overtone of this pipe has same frequency as first overtone of a closed organ pipe. If speed of sound is 330 m/s, then the length of closed organ pipe is

41 cm
30 cm
45 cm
35 cm

Solution:

(a) For open pipe,

n_1 = \frac{v}{2l}

, where

n_1

is the fundamental frequency of open pipe,

length of open pipe is.

∴ l =

\frac{v}{2n} = \frac{330}{2 \times 300}

=

\frac{11}{20}

1st overtone of open pipe,

n_2 = 2n_1

=

2\left(\frac{v}{2l}\right)

1st overtone of closed pipe,

n_3 = 3n_1

=

3\left(\frac{v}{4l'}\right)

where,

l'

= length of closed pipe

As freq. of 1 st overtone of open pipe = freq.

of 1st overtone of closed pipe

∴

2 \frac{v}{2l}

=

3 \frac{v}{4l'}

⇒

l'

=

\frac{3l}{4} = \frac{3}{4} \times \frac{11}{20}

= 41.25 cm

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