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(C) The speed of a transverse pulse on a string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density

Vedclass · Accepted Answer

$6.8$

Vedclass · Answer

(C) The speed of a transverse pulse on a string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density of the string.When the car is at rest, the tension in the string is $T_1 = Mg$. Thus, $v_1 = \sqrt{\frac{Mg}{\mu}} = 60 	ext{ cm/s}$.When the car accelerates with acceleration $a$ on a horizontal road, the effective acceleration acting on the ball is $g_{eff} = \sqrt{a^2 + g^2}$. The tension in the string becomes $T_2 = M\sqrt{a^2 + g^2}$.Thus, $v_2 = \sqrt{\frac{M\sqrt{a^2 + g^2}}{\mu}} = 66 	ext{ cm/s}$.Dividing the two equations:$\frac{v_2}{v_1} = \sqrt{\frac{\sqrt{a^2 + g^2}}{g}} = \frac{66}{60} = 1.1$.Squaring both sides:$\frac{\sqrt{a^2 + g^2}}{g} = (1.1)^2 = 1.21$.$\sqrt{a^2 + g^2} = 1.21g$.Squaring again:$a^2 + g^2 = (1.21)^2 g^2 = 1.4641 g^2$.$a^2 = 0.4641 g^2$.$a = \sqrt{0.4641} 	imes g = 0.68125 	imes 10 	ext{ m/s}^2 \approx 6.8 	ext{ m/s}^2$.

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