Question

A toy cart is tied to the end of an un-stretched string of length $l$. When revolved, the toy card moves in horizontal circle with radius '$2l$' and time period T. If it is speeded until it moves in horizontal circle of radius $3l$ with period ${T_1}$​, relation between T and ${T_1}$​ is
(A) ${T_1} = \dfrac{2}{{\sqrt 3 }}T$
(B) ${T_1} = \sqrt {\dfrac{3}{2}} T$
(C) ${T_1} = \sqrt {\dfrac{2}{3}} T$
(D) ${T_1} = \dfrac{{\sqrt 3 }}{2}T$

Answer 1

Hint: In order to solve this question, we will equate the centripetal force acting on the toy with the restoring force acting on the string due to Hook’s law and then we will solve for the time period in both the cases as given in the question.

Formula used:
1. Centripetal force acting on a body moving in a circular path is given by:
$F = m{\omega ^2}r$
 where, m is the mass, r is the radius and $\omega = \dfrac{{2\pi }}{T}$ is the angular frequency with time period T
2. Hooke’s law gives the restoring force for a string of original length l with its restoring force constant k as $F = kl$

Complete answer:
According to the question, when the time period of the toy is T then the radius of the circular path is $r = 2l$ and let ‘m’ be the mass of the toy then equating restoring force and centripetal force we get,
$
  m{\omega ^2}r = kl \\
  m{\omega ^2}(2l) = kl \\
  m{(\dfrac{{2\pi }}{T})^2}2 = k \to (i) \\
 $

Now, when the time period is ${T_1}$ the radius of the circular path is $r = 3l$ but for restoring force the length is $2l$ because it was the original length of the string when it gets stretched to $r = 3l$ hence, again equating both forces we get,
$
  m{\omega _1}^2(3l) = k(2l) \\
  m{(\dfrac{{2\pi }}{{{T_1}}})^2}3 = 2k \to (ii) \\
 $

Now, divide the equations (i) by (ii) we get,
$
  \dfrac{{{T_1}^2}}{{{T^2}}} = \dfrac{3}{4} \\
  {T_1} = \dfrac{{\sqrt 3 }}{2}T \\
 $

Hence, the correct answer is option (D) ${T_1} = \dfrac{{\sqrt 3 }}{2}T$

Note: It should be remembered that, while calculating restoring force as $F = kl$ ‘l’ is the original length of the string before stretched as in the second case the elongation of the string is ‘l’ from $2l \to 3l$ hence, always take care of lengths in such calculations to avoid any minor error.