A solid ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at the speed $60 \mathrm{~cm}^{-1}$ on the string, when the car is at rest. When the car accelerates on a horizontal road, then speed of the pulse is $66 \mathrm{~cm}^{-1}$. The…

Question

A solid ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at the speed $60 \mathrm{~cm}^{-1}$ on the string, when the car is at rest. When the car accelerates on a horizontal road, then speed of the pulse is $66 \mathrm{~cm}^{-1}$. The acceleration of the car is nearly $\left(g=10 \mathrm{~ms}^{-2}\right)$

Aicharya · Accepted Answer

When car is at rest, tension in string is $T=m g$. $ \begin{aligned} T & =M g=m v_1^2 \\ V_1=\sqrt{\frac{T}{r}} & =\sqrt{\frac{M g}{\mu}} &..(i) \end{aligned} $

When car is accelerating tension, $ \begin{aligned} T & =\sqrt{M\left(a^2+g^2\right)^{1 / 2}} \\ \therefore \quad v_2 & =\sqrt{\frac{M\left(a^2+g^2\right)^{1 / 2}}{\mu}} \end{aligned} $ Dividing Eq (i) by Eq. (ii), we get $ \begin{gathered} \frac{v_2}{v_1}=\frac{\sqrt{M\left(a^2+g^2\right)^{1 / 2}}}{\sqrt{M g}}=\frac{66}{60} \\ \frac{\left(a^2+g^2\right)^{1 / 2}}{g}=\frac{121}{100} \end{gathered} $ Squaring, we get $ \begin{array}{rlrl} & \Rightarrow & a^2+g^2 & =\left(\frac{121}{100}\right)^2 \\ \Rightarrow & a^2 & =146.41-100 \\ \Rightarrow & a^2 & =46.41 \end{array} $ So, $a=6.8 \mathrm{~ms}^{-2}$

A solid ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels…

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