# A simple pendulum of length l suspended inside a trolly which is coming on an inclined plane of inclination theta the time period is

1
by 2092000

Log in to add a comment

by 2092000

Log in to add a comment

The Brainliest Answer!

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

If we want to do this in the simplest way, the equilibrium position for the pendulum is given by angle α in the direction opposite to movement of the trolley car. Pseudo force m g sin α acts on the bob up the incline. It cancels the component of weight of bob mg sin α down the plane.

In the equilibrium position, the 3 forces, m g , Tension, and mg Sin α, result in : T = mg cos α. The component of gravity acting on the bob, along the string in equilibrium position is : g’ = g cos α.

Simply substitute g’ for g in the standard formula for time period of simple pendulum.

answer: T = 2 π √[ L / g' ]

========================

If the trolley car is moving with a uniform speed down the plane, then time period is as usual: T =

Let the trolley car is freely moving from rest, without friction an acceleration of (g Sin α).

We have the pendulum of mass m inside an accelerating vehicle. The pendulum experiences a pseudo force of m g Sin α in the opposite direction ie., upwards along the slope.

balancing forces in the vertical direction =>

T Cos Ф + mg Sin² α = mg

T Cos Ф = m g Cos² α --- -(1)

T = m g Cos²α / Cos Ф --- (2)

balancing forces in the horizontal direction

T Sin Ф = m g Sin α Cos α --- (3)

(3) / (1) => Tan Ф = tan α

=> Ф = α

So the pendulum is in equilibrium at the angle of the incline = α.

=====

Suppose from equilibrium position α, the pendulum bob is deflected by an angle β to one side. At this position, the 3 forces on the bob are: m g sin α at angle α with the horizontal upwards long the plane. Force mg is vertically downwards. Tension T in the string with an angle α + β with the vertical and hence, the tangent to the circular arc makes an angle 90 - (β+α) with the vertical.

Angle between the tangent to the circular arc and vertical direction is 90 - α.

Angle between the pseudo force m g Sin α and the tangential direction

= 90 - α - 90 + β + α = β

Hence, the angle between pseudo force and string (or tension T) = 90-β

Forces along the thread are balanced.

T = m g Cos (α+β) + m g Sin α * Cos (90 - β)

= mg [ Cos (α+β) + Sin α Sinβ ] = m g Cosα Cos β

Resultant of the forces along the tangential direction is:

F = m g Sin (α+β) - m g Sin α Cosβ = mg Cos α Sin β

But F = restoration force = m a = - m L d² β / d t²

Hence - m d² β / d t² = m g Cos α Sin β

d² β / d t² ≈ - (g Cos α / L) β approximately for small β

So the pendulum oscillates in a SHM with an angular frequency ω.

ω² = g Cos α / L

ω = √{g Cos α/L)

Time period = T = 2 π √ [L / (g Cosα) ]

T = 2 π √[ L / g' ] where g' = g Cos α

The value of tension in the string is also multiplied by a factor (cos α) compared to the regular simple pendulum.