A 20 cm long string,having a mass of 1.0 g, | vedclass

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(A) The velocity of a wave in a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the string and $\mu$ is the mass per unit le

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$5$

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(A) The velocity of a wave in a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the string and $\mu$ is the mass per unit length of the string.Given: $T = 0.5 \ N$,mass $m = 1.0 \ g = 1.0 	imes 10^{-3} \ kg$,and length $L = 20 \ cm = 0.2 \ m$.The linear mass density $\mu = \frac{m}{L} = \frac{1.0 	imes 10^{-3} \ kg}{0.2 \ m} = 0.5 	imes 10^{-2} \ kg/m$.Substituting these values into the velocity formula:$v = \sqrt{\frac{0.5}{0.5 	imes 10^{-2}}} = \sqrt{100} = 10 \ m/s$.The wavelength $\lambda$ is given by $\lambda = \frac{v}{f}$,where $f = 100 \ Hz$.$\lambda = \frac{10 \ m/s}{100 \ Hz} = 0.1 \ m = 10 \ cm$.The separation between successive nodes in a stationary wave is $\frac{\lambda}{2}$.Therefore,the separation $= \frac{10 \ cm}{2} = 5 \ cm$.

$A$ $20 \ cm$ long string,having a mass of $1.0 \ g$,is fixed at both the ends. The tension in the string is $0.5 \ N$. The string is set into vibrations using an external vibrator of frequency $100 \ Hz$. Find the separation (in $cm$) between the successive nodes on the string.

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