The equation of a wave on a string of liner mass density 0.04 kg$m^{- 1}$is given by $y = 0.02(m)\sin\left\lbrack 2\pi\left( \frac{t}{0.04(s)} - \frac{x}{0.50(m)} \right) \right\rbrack$. the tension in the string is

Question

Dr. Abubacker · Accepted Answer

(d) $T = \mu v^{2} = \mu\frac{\omega^{2}}{k^{2}} = 0.04\frac{\left( \frac{2\pi}{0.004} \right)^{2}}{\left( \frac{2\pi}{0.50} \right)^{2}} = 6.25N$

Q. The equation of a wave on a string of liner mass density 0.04 kg$m^{- 1}$is given by $y = 0.02(m)\sin\left\lbrack 2\pi\left( \frac{t}{0.04(s)} - \frac{x}{0.50(m)} \right) \right\rbrack$. the tension in the string is

Explanation

Reviewed by Dr. Abubacker

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