Waves

JEE Main level, single correct, medium

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Question 18 of 142

Equation of travelling wave on a stretched string of linear density 5 g/m5 \mathrm{~g} / \mathrm{m} is y=0.03sin⁡(450t−9x)\mathrm{y}=0.03 \sin (450 \mathrm{t}-9 \mathrm{x}) where distance and time are measured in SI units. The tension in the string is:

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Answer: (C) 12.5 N12.5 \mathrm{~N}

We have given, y=0.03sin⁡(450t−9x)y=0.03 \sin (450 t-9 x)

Comparing it with standard equation of wave, we get ω=\omega=

450k=9450 \mathrm{k}=9

∴v=ωk=4509=50 m/s\therefore \mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{450}{9}=50 \mathrm{~m} / \mathrm{s}

Velocity of travelling wave on a stretched string is given by v=Tμ⇒Tμ=2500\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}} \Rightarrow \frac{\mathrm{T}}{\mu}=2500

μ=\mu= linear mass density ⇒T=2500×5×10−3\Rightarrow \mathrm{T}=2500 \times 5 \times 10^{-3}

⇒12.5 N\Rightarrow 12.5 \mathrm{~N}

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