Wave Motion Ques 99
- A string $25 cm$ long and having a mass of $2.5 g$ is under tension. A pipe closed at one end is $40 cm$ long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats/s are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is $320 m / s$ find the tension in the string.
(1982, 7M)
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Answer:
Correct Answer: 99.27.04 N
Solution:
Formula:
- By decreasing the tension in the string beat frequency is decreasing, it means frequency of string was greater than frequency of pipe. Thus,
First overtone frequency of string - Fundamental frequency of closed pipe $=8$
$\therefore 2\left(\frac{v _1}{2 l _1}\right)-\left(\frac{v _2}{4 l _2}\right)=8$ or $v _1=l _1\left[8+\frac{v _2}{4 l _2}\right]$
Substituting the value, we have
$$ \begin{aligned} v _1 & =0.25\left[8+\frac{320}{4 \times 0.4}\right]=52 m / s \\ \text { Now, } \quad v _1 & =\sqrt{\frac{T}{\mu}} \\ \therefore \quad T & =\mu v _1^{2}=\left(\frac{m}{l}\right) v _1^{2}=\left(\frac{2.5 \times 10^{-3}}{0.25}\right)(52)^{2}=27.04 N \end{aligned} $$
BETA
