JEE Mains · Physics · STD 11 - 14. waves and sound
Equation of travelling wave on a stretched string of linear density \(5\,g/m\) is \(y = 0.03\,sin\,(450\,t -9x)\) where distance and time are measured in \(SI\) united. The tension in the string is ... \(N\)
- A \(10\)
- B \(7.5\)
- C \(12.5\)
- D \(5\)
Answer & Solution
Correct Answer
(C) \(12.5\)
Step-by-step Solution
Detailed explanation
\(y=0.03 \sin \left[450\left(t-\frac{9 x}{450}\right)\right]\) So, \(\mathrm{v}=\frac{450}{9}=50 \mathrm{m} / \mathrm{s}\) Also, \(v=\sqrt{\frac{T}{\lambda}}\) \(\Rightarrow 50=\sqrt{\frac{T}{5 \times 10^{-3}}}\) \(\Rightarrow T=2500 \times 5 \times 10^{-3}=12.5 \mathrm{N}\)
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