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,

\[\hspace{0.167em} \text{ Wave  speed,}   \nu = \sqrt{\left( \frac{T}{m} \right)} = \sqrt{\left( \frac{Mg}{m} \right)}\] 

\[\text{ When  car  is  at  rest: }\] 

\[\text{ Tension  in  the  string,}   T = Mg\] 

\[ \Rightarrow  v_1  = \sqrt{\left( \frac{Mg}{m} \right)}\] 

\[ \Rightarrow \frac{Mg}{m} =  \left( 60 \right)^2          .  .  . (i)\] 

When car is having acceleration:
Tension, 

\[T = \sqrt{\left( Ma \right)^2 + \left( Mg \right)^2}\]

\[Again,    \nu_2  = \sqrt{\left( \frac{T}{m} \right)}\] 

\[ \Rightarrow 62 = \frac{\left[ \left( Ma \right)^2 + \left( Mg \right)^2 \right]^{1/4}}{m^{1/2}}\] 
\[\Rightarrow   \sqrt{\frac{\left[ \left( Ma \right)^2 + \left( Mg \right)^2 \right]}{m}} =  \left( 62 \right)^2                .  .  . (ii)\] 
From equations (i) and (ii), we get: 

\[\left( \frac{Mg}{m} \right) \times \frac{m}{\sqrt{\left( Ma \right)^2 + \left( Mg \right)^2}} =  \left( \frac{60}{62} \right)^2 \] 

\[ \Rightarrow \frac{g}{\sqrt{\left( a^2 + g^2 \right)}} = 0 . 936\] 

\[ \Rightarrow \frac{g^2}{\left( a^2 + g^2 \right)} = 0 . 876\] 

\[ \Rightarrow \left( a^2 + 100 \right)  0 . 876 = 100\] 

\[ \Rightarrow    a^2  = \frac{12 . 4}{0 . 867} = 14 . 15\] 

\[ \Rightarrow a = 3 . 76  m/ s^2 \] 

\[\text{ Therefore,   acceleration  of  the  car  is  3 . 76  m/ s^2  .}\]