Question

A block of mass $\mathrm{M}$ is hanging by a string of negligible mass in a car. The speed of the wave in the string $60{\mathrm{ms}}^{-1}$. Now the car is accelerated horizontally by an acceleration $\mathrm{a}$ and the speed of wave in the string is $60.5{\mathrm{ms}}^{-1}$. What is $\mathrm{a}$ in terms of $\mathrm{g}$?

• A. $\frac{\mathrm{g}}{5}$
• B. $\frac{\mathrm{g}}{10}$
• C. $\frac{\mathrm{g}}{20}$
• D. $\frac{\mathrm{g}}{30}$

Answer 1

The correct option is A

$\frac{\mathrm{g}}{5}$

Step 1: Given data

The mass of a block$=\mathrm{M}$

The speed of wave in the string when car is at rest, $\mathrm{v}=60{\mathrm{ms}}^{-1}$

When car accelerates, the speed of wave$=60.5{\mathrm{ms}}^{-1}$

Mass of spring is negligible

Step 2: Finding the value of $\mathrm{a}$ in terms of $\mathrm{g}$

The formula we will use here is,

$\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mathrm{\mu }}}$

$\mathrm{v}\Rightarrow$velocity of wave

$\mathrm{T}\Rightarrow$tension in the string$=\mathrm{Mg}$

$\mathrm{\mu }\Rightarrow$linear density of a string

Hence, the wave velocity when the car is at rest,

$$\begin{gathered} \mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mathrm{\mu }}} \\ \Rightarrow 60{\mathrm{ms}}^{-1}=\sqrt{\frac{\mathrm{Mg}}{\mathrm{\mu }}}\left(1\right) \end{gathered}$$

When car will accelerate, the new value of $\mathrm{g}$ will be,

$\mathrm{g}\text{'}=\sqrt{{\mathrm{a}}^2+{\mathrm{g}}^2}$

Hence, the velocity of wave in this case will be,

$$\begin{gathered} \mathrm{v}=\sqrt{\frac{\mathrm{M}\sqrt{{\mathrm{a}}^2+{\mathrm{g}}^2}}{\mathrm{\mu }}} \\ \Rightarrow 60.5{\mathrm{ms}}^{-1}=\sqrt{\frac{\mathrm{M}\sqrt{{\mathrm{a}}^2+{\mathrm{g}}^2}}{\mathrm{\mu }}}\left(2\right) \end{gathered}$$

Divide $\left(2\right)$ by $\left(1\right)$ we get,

$$\begin{gathered} \frac{60.5{\mathrm{ms}}^{-1}}{60{\mathrm{ms}}^{-1}}=\frac{{\displaystyle \frac{\sqrt{\mathrm{M}\sqrt{{\mathrm{a}}^2+{\mathrm{g}}^2}}}{\mathrm{\mu }}}}{{\displaystyle \frac{\sqrt{\mathrm{Mg}}}{\mathrm{\mu }}}} \\ \Rightarrow \frac{60.5}{60}=\frac{\sqrt{\sqrt{{\mathrm{a}}^2+{\mathrm{g}}^2}}}{\sqrt{\mathrm{g}}} \end{gathered}$$

Squaring on both sides we get,

${\left(\frac{60.5}{60}\right)}^2=\frac{\sqrt{{\mathrm{a}}^2+{\mathrm{g}}^2}}{\mathrm{g}}$

Again squaring on both sides we get,

$$\begin{gathered} {\left(\frac{60.5}{60}\right)}^4=\frac{{\mathrm{a}}^2+{\mathrm{g}}^2}{{\mathrm{g}}^2} \\ \Rightarrow {\mathrm{a}}^2={\mathrm{g}}^2\times {\left(\frac{60.5}{60}\right)}^4-{\mathrm{g}}^2 \\ \Rightarrow \mathrm{a}=\mathrm{g}\sqrt{{\left(\frac{60.5}{60}\right)}^4-1} \\ \Rightarrow \mathrm{a}=0.1837\mathrm{g} \\ \Rightarrow \mathrm{a}=\frac{\mathrm{g}}{5} \end{gathered}$$

Therefore the value of $\mathrm{a}$ in terms of $\mathrm{g}$ is equal to $\frac{\mathrm{g}}{5}$

Hence, option(A) is the correct answer.