A hollow pipe of length 0.8m is closed a 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 50N and the speed of sound $320\ ms^{- 1}$, the mass of the string is

Question

Dr. Abubacker · Accepted Answer

(b) $2\left( \frac{v_{1}}{2\iota_{1}} ight) = \frac{v_{2}}{4\iota_{2}}$
$	herefore\frac{\sqrt{T/\mu}}{\iota_{1}} = \frac{320}{4\iota_{2}}\ $
($\mu$=mass per unit length of wire)
$Or\ \frac{\sqrt{50/\mu}}{0.5} = \frac{320}{4 	imes 0.8}$
Solving we get $\mu$=0.02 kg/m=20 g/m
∴ Mass of string=20 g/m ×0.5 m=10 g

Q. A hollow pipe of length 0.8m is closed a 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 50N and the speed of sound $320\ ms^{- 1}$, the mass of the string is

Explanation

Reviewed by Dr. Abubacker

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