Waves

JEE Main level, single correct, medium

77 of 142

Question 77 of 142

The equation of a wave on a string of linear mass density 0.04 kg m−10.04 \mathrm{~kg} \mathrm{~m}^{-1} is given by y=0.02( m)sin⁡[2π(t0.04( s)−x0.50( m))]y=0.02(\mathrm{~m}) \sin \left[2 \pi\left(\frac{\mathrm{t}}{0.04(\mathrm{~s})}-\frac{\mathrm{x}}{0.50(\mathrm{~m})}\right)\right]. The tension in the string is

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Answer: (D) 6.25 N6.25 \mathrm{~N}

T=μv2=μω2k2=0.04(2π/0.004)2(2π/0.50)2=6.25 N\mathrm{T}=\mu \mathrm{v}^2=\mu \frac{\omega^2}{\mathrm{k}^2}=0.04 \frac{(2 \pi / 0.004)^2}{(2 \pi / 0.50)^2}=6.25 \mathrm{~N}

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