Question

Answer the following question.

Show that its time period is given by, 2π`sqrt((l cos theta)/("g"))` where l is the length of the string, θ is the angle that the string makes with the vertical, and g is the acceleration due to gravity.

Answer 1

Conical pendulum

Where,
O: rigid support,
T: tension in the string,
l: length of string,
h: height of support from bob,
v: velocity of bob,
r: radius of horizontal circle,
θ: semi-vertical angle,
mg: weight of bob

1. Consider a bob of mass m tied to one end of a string of length ‘l’ and the other end is fixed to a rigid support.
2. Let the bob be displaced from its mean position and whirled around a horizontal circle of radius ‘r’ with constant angular velocity ω, then the bob performs U.C.M.
3. During the motion, string is inclined to the vertical at an angle θ as shown in the figure above.
4. In the displaced position, there are two forces acting on the bob.
   a. The weight mg acting vertically downwards.
   b. The tension T acting upward along the string.
5. The tension (T) acting in the string can be resolved into two components:
   a. T cos θ acting vertically upwards.
   b. T sin θ acting horizontally towards centre of the circle.
6. Since there is no net force, the vertical component T cos θ balances the weight and the horizontal component T sin θ provides the necessary centripetal force.
   ∴ T cos θ = mg      ....(1)
   T sin θ = `"mv"^2/"r" = "mr"omega^2`   ....(2)
7. Dividing equation (2) by (1),
   tan θ = `"v"^2/"rg"`      ....(3)
   Therefore, the angle made by the string with the vertical is θ = tan^-1 `("v"^2/"rg")`
8. Since we know v = `(2pi"r")/"T"`
   ∴ tan θ = `(4pi^2"r"^2)/("T"^2"rg")`     ....[From (3)]
   T = `2pi sqrt("r"/("g"tan theta))`
   T = `2pi sqrt((l sin theta)/("g"tan theta))  ....[because "r" = l sin theta]`
   T = `2pi sqrt((l cos theta)/("g"))`
   T = `2pi sqrt("h"/"g")`            .....(∵ h = l cos θ)

where l is length of the pendulum and h is the vertical distance of the horizontal circle from the fixed point O.