The equation of a wave on a string of linear mass density 0.04 kgm−1 is given by Y=0.02(m) sin[2π(t0.04(sec)−x0.5(m))] . The tension in the string is

v=Tμ=ωk= coefficient of 't' coefficient of 'x'T0.04=2π/0.042π/0.5=0.50.04=252T=(252)2×0.04T = 6.25 N

The equation of a wave on a string of linear mass density 0.04 kgm−1 is given by Y=0.02(m) sin[2π(t0.04(sec)−x0.5(m))] . The tension in the string is

v=Tμ=ωk= coefficient of 't' coefficient of 'x'T0.04=2π/0.042π/0.5=0.50.04=252T=(252)2×0.04T = 6.25 N

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The equation of a wave on a string of linear mass density $0.04 k g m^{- 1}$ is given by $Y = 0.02 (m) \sin [2 π (\frac{t}{0.04 (\sec)} - \frac{x}{0.5 (m)})]$ . The tension in the string is

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12.5 N

0.5 N

4.0 N

6.25 N

answer is D.

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$v = \sqrt{\frac{T}{μ}} = \frac{ω}{k} = \frac{c o e f f i c i e n t o f' t'}{c o e f f i c i e n t o f' x'}$

$\sqrt{\frac{T}{0.04}} = \frac{2 π / 0.04}{2 π / 0.5} = \frac{0.5}{0.04} = \frac{25}{2}$

$T = {(\frac{25}{2})}^{2} \times 0.04$

$T = 6.25 N$

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The equation of a wave on a string of linear mass density 0.04 kgm−1 is given by Y=0.02(m) sin[2π(t0.04(sec)−x0.5(m))] . The tension in the string is

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The equation of a wave on a string of linear mass density $0.04 k g m^{- 1}$ is given by $Y = 0.02 (m) \sin [2 π (\frac{t}{0.04 (\sec)} - \frac{x}{0.5 (m)})]$ . The tension in the string is