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Section: Physics

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Question : 25 of 31

Marks: +1, -0

A hollow pipe of length 0.8 m is closed at one end. At its open end, a 0.5-m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320

\text{m/s}

, the mass of the string is

[JEE Adv 2010 P2]

5 g
10 g
20 g
40 g

Solution:

The fundamental frequency of the pipe is

\frac{v}{4l}

. The second harmonic of vibrations in string is

\frac{2}{2L} \sqrt{\frac{T}{\mu}}

=

\frac{1}{L} \sqrt{\frac{TL}{m}}

where μ is mass per length of the string and m is the mass of the string. Therefore,

\frac{v}{4l}

=

\frac{1}{L} \sqrt{\frac{TL}{m}}

⇒

\frac{v^2}{16l^2}

=

\frac{1}{L} \left( \frac{T}{m} \right)

⇒ m =

\frac{16 l^2 T}{L v^2}

=

\frac{16 \times 0.64 \times 50}{0.5 \times 320 \times 320}

= 0.01 kg = 10 g

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