Wave Motion Ques 66
- A uniform rope of length $12 m$ and mass $6 kg$ hangs vertically from a rigid support. A block of mass $2 kg$ is attached to the
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Answer:
Correct Answer: 66.$0.12 m$
Solution:
Formula:
Speed of Transverse Wave Along a String/Wire:
- $v=\sqrt{T / \mu}$
$$ \frac{v _{\text {top }}}{v _{\text {bottom }}}=\sqrt{\frac{T _{\text {top }}}{T _{\text {bottom }}}}=\sqrt{\frac{6+2}{2}}=2 $$
Frequency will remain unchanged. Therefore, Eq. (i) can be written as,
$$ \begin{gathered} \frac{f \lambda _{\text {top }}}{f \lambda _{\text {bottom }}}=2 \\ \text { or } \quad \lambda _{\text {top }}=2\left(\lambda _{\text {bottom }}\right)=2 \times 0.06=0.12 m \end{gathered} $$
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