Wave Motion Ques 23
- 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)
(a) $5 g$
(b) $10 g$
(c) $20 g$
(d) $40 g$
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Answer:
Correct Answer: 23.(b)
Solution:
Formula:
Given, 2nd harmonic of I = Fundamental of II
$$ \begin{array}{rlrl} & \therefore \quad 2\left(\frac{v _1}{2 l _1}\right) & =\frac{v _2}{4 l _2} \Rightarrow \frac{T / \mu}{l _1}=\frac{v _2}{4 l _2} \\ \Rightarrow & \mu & =\frac{16 T l _2^{2}}{v _2^{2} l _1^{2}}=\frac{16 \times 50 \times(0.8)^{2}}{(320)^{2} \times(0.5)^{2}} \\ & =0.02 kg / m \\ & \therefore \quad m _1=\mu l _1 & =(0.02)(0.2)=0.01 kg=10 g \end{array} $$
BETA
