Question

Minimum time period in a compound pendulum is obtained when
A. $l = \pm \dfrac{K}{2}$
B. $l = \pm K$
C. $l = \pm \dfrac{K}{{\sqrt 2 }}$
D. $l = 0$

Answer 1

Hint- In order to find the value at which time period becomes minimum we can take the derivative of the equation for time period and equate it to zero. By solving that equation we can reach the condition at which time period becomes zero.

Step by step solution:
The time period of a compound pendulum is given as
$T = 2{\left( {\dfrac{I}{{mgh}}} \right)^{\dfrac{1}{2}}}$
Where, I is the moment of inertia of the compound pendulum, m is the mass, g is the acceleration due to gravity and h is the height.
The moment of inertia about the centre of mass is the product of mass and the square of radius of gyration.
${I_c} = m{K^2}$
The moment of inertia is calculated using the parallel axis theorem. The moment of inertia about an Axis passing through one end of the compound pendulum will be the sum of moment of inertia of an Axis passing through the centre of mass and the product of mass and square of distance between the two parallel axes.
$\Rightarrow I = {I_c} + m{h^2}$
$\Rightarrow I = m{K^2} + m{h^2}$
Thus, the time period can be written as
$\Rightarrow T = 2{\left( {\dfrac{{m{K^2} + m{h^2}}}{{mgh}}} \right)^{\dfrac{1}{2}}}$
In order to find the minimum, we can take the differential of this equation and equate it to zero.
$\Rightarrow dT = 0$
$\Rightarrow \dfrac{1}{2}{\left( {\dfrac{{m{K^2} + m{h^2}}}{{mgh}}} \right)^{ - \dfrac{1}{2}}}\dfrac{{mgh(2mh) - mg(m{K^2} + m{h^2})}}{{{{(mgh)}^2}}} = 0$
$\Rightarrow mgh(2mh) - mg(m{K^2} + m{h^2}) = 0$
$\Rightarrow 2{h^2} - {h^2} - {K^2} = 0$
$\therefore h = \pm K$
The value of h in this case is the distance from the hinge point to the centre of mass. It is represented as l.
$\therefore l = \pm K$
This is the value for which we get the minimum time period.

So, the correct answer is option B.

Note: Remember that the moment of inertia depends upon the axis of rotation. In the case of a compound pendulum the axis is at the hinge point, so to find the moment of inertia about this axis we can consider an axis passing through the centre of mass of the pendulum parallel to this axis. Then we can use the parallel axis theorem to find the moment of inertia.