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Q. 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 $50\,ms^{- 1}$ when the car is at rest. But if the car is imparted an acceleration of $a\,ms^{- 2}$ , the wave speed increases to $51\,ms^{- 1}$ . [Take $g = 10 \,m s^{- 2}]$ . Write the value of $\left[a\right],$ where $\left[\right]$ is the greatest integer function.

NTA AbhyasNTA Abhyas 2022

Solution:

Case I:
Solution
From diagram, $T_{1}=Mg$
$\therefore v_{1}=\sqrt{\frac{T_{1}}{\mu }}=\sqrt{\frac{M g}{\mu }}$ ....(i)
Case II:
Solution
From diagram, $T_{2}cos\theta =Mg,T_{2}sin\theta =Ma$
$\therefore T_{2}^{2}\left(\left(sin\right)^{2} \theta + \left(cos\right)^{2} \theta \right)=M^{2}\left(g^{2} + a^{2}\right)$
$\Rightarrow T_{2}=M\left(g^{2} + a^{2}\right)^{\frac{1}{2}}$
$v_{2}=\sqrt{\frac{T_{2}}{\mu }}=\sqrt{\frac{M \sqrt{a^{2} + g^{2}}}{\mu }}....\left(i i\right)$
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 $\frac{v_{2}}{v_{1}}=1+\frac{1}{4}\frac{a^{2}}{g^{2}}\ldots .$
$\left[\because \left( 1 + x \right)^{n} \approx 1 + nx\right]$
$\Rightarrow \frac{51}{50}=1+\frac{1}{4}\frac{a^{2}}{g^{2}}$
$\Rightarrow \frac{a^{2}}{g^{2}}=\frac{8}{100}$
$\Rightarrow a=\frac{2 \sqrt{2} g}{10}$
$\Rightarrow a=2.83\,ms^{- 2}$
$\left[a\right]=2$