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 3

Question

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 ${ms}^{- 1}$ , the mass of the string is [2010]

Smart Achievers · Accepted Answer

(2)

$Frequency of 2nd harmonic of string = fundamental frequency produced in the pipe$

$2 (\frac{v_{1}}{2 l_{1}}) = \frac{v_{2}}{4 l_{2}}$

$∴$ $2 \times [\frac{1}{2 l_{1}} \sqrt{\frac{T}{μ}}] = \frac{v}{4 l_{2}}$

$∴$ $\frac{1}{0.5} \sqrt{\frac{50}{μ}} = \frac{320}{4 \times 0.8}$ $\Rightarrow μ = 0.02 {kg m}^{- 1}$

Hence, the mass of the string, $m_{1} = μ l_{1}$

$= 0.02 \times 0.5 kg = 10 g$

Solution

1 5 grams

2 10 grams

3 20 grams

4 40 grams

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