# The time period of simple pendulum hung from a length L thread from roof of a car which is going down a fixed inclined plane of angle α without friction is ?

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by namku

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by namku

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see diagram.

we assume that the car is freely moving from rest. There is no friction. The car is going down the slope with a n acceleration of g Sin α.

So inside an accelerating body, we have the pendulum. The pendulum experiences a pseudo force of m g Sin α in the opposite direction ie., upwards along the slope. We are looking at the pendulum from the reference frame stationary wrt the car.

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

=>

So the pendulum is in equilibrium at an angle = angle of inclined plane = α.

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Suppose from this mean position, the pendulum bob is deflected by an angle β to one side. At this position, the forces on the bob are :

m g sin α at angle α with the horizontal. force mg vertically downwards.

we use that Ф = α.

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.

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

The 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)

= m g Cos (α+β) + m g Sin α Sin 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 = 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 Cosα / 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) ]

Hence the pendulum oscillates with a time period slightly modified in the formula.

Also the tension in the string is also multiplied by a factor cos α compared to the regular simple pendulum.