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  I am using symbol α for the angle of inclined plane.
    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' ]
   ,        where g' = g Cos α

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Derivation is done in two steps.  Equilibrium position and then formula for SHM. See the diagram.

If the trolley car is moving with a uniform speed down the plane, then time period is as usual:             T =  2 π √[ L / g ]
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.

Equilibrium  position OP (equilibrium position) when the pendulum is not oscillating in the accelerating car:    Let the angle of the string OP with the vertical OS be = Ф.

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.

=====
 SH M of the pendulum :

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.

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