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(C) The speed of a transverse wave in a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.W

Vedclass · Accepted Answer

$\frac{g}{5}$

Vedclass · Answer

(C) The speed of a transverse wave in a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.When the car is at rest,the tension $T_1 = Mg$. Thus,$60 = \sqrt{\frac{Mg}{\mu}}$.When the car accelerates with $a$,the effective acceleration is $g_{eff} = \sqrt{g^2 + a^2}$. The tension becomes $T_2 = M\sqrt{g^2 + a^2}$.Thus,$60.5 = \sqrt{\frac{M\sqrt{g^2 + a^2}}{\mu}}$.Dividing the two equations: $\frac{60.5}{60} = \sqrt{\frac{\sqrt{g^2 + a^2}}{g}} = \left(\frac{g^2 + a^2}{g^2}ight)^{1/4}$.Raising both sides to the power of $4$: $\left(1 + \frac{0.5}{60}ight)^4 = 1 + \frac{a^2}{g^2}$.Using the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$: $1 + 4 	imes \frac{0.5}{60} = 1 + \frac{a^2}{g^2}$.$\frac{2}{60} = \frac{a^2}{g^2} \Rightarrow \frac{a^2}{g^2} = \frac{1}{30}$.$a = \frac{g}{\sqrt{30}} \approx \frac{g}{5.47} \approx \frac{g}{5}$.

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