A) 4.0 N
B) 12.5 N
C) 0.5 N
D) 6.25 N
Correct Answer: D
Solution :
[d] The given equation of a wave is \[y=0.02\sin \left[ 2\pi \left( \frac{t}{0.04}-\frac{x}{0.50} \right) \right]\] Compare it with the standard wave equation \[y=A\sin (\omega t-kx)\] We get \[\omega =\frac{2\pi }{0.04}\text{rad}\,{{s}^{-1}};k=\frac{2\pi }{0.5}\text{rad}\,{{m}^{-1}}\] Wave velocity,\[v=\frac{\omega }{k}=\frac{\left( 2\pi /0.04 \right)}{\left( 2\pi /0.5 \right)}m{{s}^{-1}}\] ...(i) Also\[v=\sqrt{\frac{T}{\mu }}\] ...(ii) Where T is the tension in the string and \[\mu \]. is the linear mass density here, linear mass density.\[\mu =0.04kg\,{{m}^{-1}}\] Equating equations (i) and (ii), we get \[\frac{\omega }{k}=\sqrt{\frac{T}{\mu }}\] or \[T=\frac{\mu {{\omega }^{2}}}{{{k}^{2}}}\] \[T=\frac{0.04\times {{\left( \frac{2\pi }{0.04} \right)}^{2}}}{{{\left( \frac{2\pi }{0.05} \right)}^{2}}}=6.25N\]
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