Question

A simple pendulum has a string of length l and bob of mass m. When the bob is at its lower position, it is given the maximum horizontal speed necessary for it to move in a circular path about the point of suspension. The tension in the string at the lowest position of the bob is
A. 6$mg$
B. $3\,mg$
C. $\sqrt {10} \,mg$
D. $4\,mg$

Answer 1

Hint In the question, length and mass are essential parameters. Here, the bob is at a lower position and it is moved along a circular path about the point of the suspension. So, by using the expression of the work-energy theorem, we get the value of the tension in the string.
Formula used:
Kinetic energy $ = \dfrac{1}{2}m{v^2}$
Potential energy $ = mg2l$
Where,
$m$ be the mass, $v$ be the velocity, $g$ be the acceleration due to gravity and $l$ be the length.

Complete step by step answer
Let A be the topmost point of the circle and B be the lowest point of the circle.
Let ${v_1}$and ${v_2}$be the velocities at A and B respectively.
Applying the principle of the conservation of the energy between A and B.
$\dfrac{1}{2}m{v_2}^2 - \dfrac{1}{2}mv_1^2 = mg2l$
Or we written the equation as
$\dfrac{{mv_2^2}}{l} = \dfrac{{mv_1^2}}{l} + 4mg..........\left( 1 \right)$
At the lowest point of the circle B, $\dfrac{{mv_2^2}}{l} = T - mg...........\left( 2 \right)$
At the top most point of the circle A, \[\dfrac{{mv_1^2}}{l} = mg...........\left( 3 \right)\]
Solving all the above three equations, we get
$mg + 4mg = T - mg$
Performing the arithmetic operation in the above equation, we get
$T = 6\,mg.$
Therefore, the tension in the lowest position of the bob is $6\,mg.$

Hence from the above options, option A is correct.

Note In the question, the pendulum is moved about a point of the suspension. So, there must be a potential and kinetic energy. By using the parameters of the work-energy theorem, we get the result.