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

by 2092000 17.09.2015

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by 2092000 17.09.2015

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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' ]

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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.

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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.