SATHEE: Wave Motion 2 Question 14

Wave Motion 2 Question 14

14. 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 $320 m s^{- 1}$ , the mass of the string is

(a) $5 g$

(b) $10 g$

(d) $40 g (2010)$

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Answer:

Correct Answer: 14. (b)

Solution:

Given, 2nd harmonic of I = Fundamental of II

$\begin{aligned} ∴ 2 (\frac{v_{1}}{2 l_{1}}) & = \frac{v_{2}}{4 l_{2}} \Rightarrow \frac{T / μ}{l_{1}} = \frac{v_{2}}{4 l_{2}} \\ \Rightarrow & μ & = \frac{16 T l_{2}^{2}}{v_{2}^{2} l_{1}^{2}} = \frac{16 \times 50 \times (0.8)^{2}}{(320)^{2} \times (0.5)^{2}} \\ = 0.02 k g / m \\ ∴ m_{1} = μ l_{1} & = (0.02) (0.2) = 0.01 k g = 10 g \end{aligned}$

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Wave Motion 2 Question 14

14. A hollow pipe of length 0.8m is closed at one end. At its open end a 0.5m 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 is 320ms−1, the mass of the string is

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