The equation of a wave on a string of linear | vedclass

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(A) The given wave equation is $Y = 0.01 \sin [2\pi (t/0.02 - x/0.50)] 	ext{ m}$.Comparing this with the standard wave equation $Y = A \sin (\omega t

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(A) The given wave equation is $Y = 0.01 \sin [2\pi (t/0.02 - x/0.50)] 	ext{ m}$.Comparing this with the standard wave equation $Y = A \sin (\omega t - kx)$, we get:$\omega = 2\pi / 0.02 = 100\pi 	ext{ rad/s}$$k = 2\pi / 0.50 = 4\pi 	ext{ m}^{-1}$The wave speed $v$ is given by $v = \omega / k = 100\pi / 4\pi = 25 	ext{ m/s}$.The speed of a wave on a string is also given by $v = \sqrt{T/\mu}$, where $T$ is the tension and $\mu$ is the linear mass density.Given $\mu = 0.02 	ext{ kg m}^{-1}$, we have:$25 = \sqrt{T / 0.02}$Squaring both sides: $625 = T / 0.02$$T = 625 	imes 0.02 = 12.5 	ext{ N}$.

The equation of a wave on a string of linear mass density $0.02 \text{ kg m}^{-1}$ is $Y = 0.01 \sin [2\pi (t/0.02 - x/0.50)] \text{ m}$. The tension in the string is (in $\text{ N}$)

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