Question

A heavy ball of mass $M$ is suspended from the ceiling of a car by a light string of mass $m\left(m<<M\right).$ The speed of transverse waves in the string is $50m{s}^{-1}$ when the car is at rest. But if the car is imparted an acceleration of $am{s}^{-2}$ , the wave speed increases to $51m{s}^{-1}$ . [Take $g=10m{s}^{-2}]$ . Write the value of $\left[a\right],$ where $\left[\right]$ is the greatest integer function.

Answer 1

Answer: 2

Case I:

From diagram, $T_1=Mg$
$\therefore v_1=\sqrt{\frac{T_1}{\mu }}=\sqrt{\frac{Mg}{\mu }}$ ....(i)
Case II:

From diagram, $T_{2}cos\theta =Mg,T_{2}sin\theta =Ma$
$\therefore T_2^2\left({\left(sin\right)}^2\theta +{\left(cos\right)}^2\theta \right)=M^2\left(g^2+a^2\right)$
$\Rightarrow T_2=M{\left(g^2+a^2\right)}^{\frac{1}{2}}$
$v_2=\sqrt{\frac{T_2}{\mu }}=\sqrt{\frac{M\sqrt{a^2+g^2}}{\mu }}....\left(ii\right)$
From (i) and (ii), $\frac{v_1}{v_2}=\sqrt{\frac{g}{{\left(g^2+a^2\right)}^{\frac{1}{2}}}}$
$\Rightarrow \left(\frac{v_2}{v_1}\right)={\left(\frac{g^2+a^2}{g^2}\right)}^{\frac{1}{4}}$
Taking binomial approximation, we get $\frac{v_2}{v_1}=1+\frac{1}{4}\frac{a^2}{g^2}\dots .$
$\left[\because {\left(1+x\right)}^n\approx 1+nx\right]$
$\Rightarrow \frac{51}{50}=1+\frac{1}{4}\frac{a^2}{g^2}$
$\Rightarrow \frac{a^2}{g^2}=\frac{8}{100}$
$\Rightarrow a=\frac{2\sqrt{2}g}{10}$
$\Rightarrow a=2.83m{s}^{-2}$
$\left[a\right]=2$