Question

A heavy ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at a speed of 60 cm s ⁻¹ on the string when the car is at rest and 62 cm s⁻¹ when the car accelerates on a horizontal road. Find the acceleration of the car. Take g = 10 m s⁻².

Answer 1

Given,
Speed of the transverse pulse when the car is at rest, v₁ = 60 cm s⁻¹
Speed of the transverse pulse when the car accelerates, v₂= 62 cm s⁻¹
Let:
Mass of the heavy ball suspended from the ceiling = M
Mass per unit length = m
Now,
$$\begin{gathered} \hspace{0.17em}\mathrm{Wave}\mathrm{speed},\nu =\sqrt{\left(\frac{T}{m}\right)}=\sqrt{\left(\frac{Mg}{m}\right)} \\ \mathrm{When}\mathrm{car}\mathrm{is}\mathrm{at}\mathrm{rest}: \\ \mathrm{Tension}\mathrm{in}\mathrm{the}\mathrm{string},T=\mathrm{M}g \\ \Rightarrow v_1=\sqrt{\left(\frac{Mg}{m}\right)} \\ \Rightarrow \frac{Mg}{m}={\left(60\right)}^2...\left(\mathrm{i}\right) \end{gathered}$$
When car is having acceleration:
Tension, $T=\sqrt{{\left(Ma\right)}^2+{\left(Mg\right)}^2}$
$$\begin{gathered} \mathrm{Again},{\nu }_2=\sqrt{\left(\frac{T}{m}\right)} \\ \Rightarrow 62=\frac{{\left[{\left(Ma\right)}^2+{\left(Mg\right)}^2\right]}^{1/4}}{m^{1/2}} \end{gathered}$$
$\Rightarrow \sqrt{\frac{\left[{\left(\mathrm{M}a\right)}^2+{\left(\mathrm{M}g\right)}^2\right]}{m}}={\left(62\right)}^2...\left(\mathrm{ii}\right)$
From equations (i) and (ii), we get:
$$\begin{gathered} \left(\frac{Mg}{m}\right)\times \frac{m}{\sqrt{{\left(Ma\right)}^2+{\left(Mg\right)}^2}}={\left(\frac{60}{62}\right)}^2 \\ \Rightarrow \frac{g}{\sqrt{\left(a^2+g^2\right)}}=0.936 \\ \Rightarrow \frac{g^2}{\left(a^2+g^2\right)}=0.876 \\ \Rightarrow \left(a^2+100\right)0.876=100 \\ \Rightarrow a^2=\frac{12.4}{0.867}=14.15 \\ \Rightarrow a=3.76\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Therefore},\mathrm{acceleration}\mathrm{of}\mathrm{the}\mathrm{car}\mathrm{is}3.76\mathrm{m}/{\mathrm{s}}^2. \end{gathered}$$