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Answer:     T  =  2 π √[ L / g' ]   ,        where g' = g Cos α
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.

Mean 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 an angle = angle of inclined plane = α.

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) ]

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

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.

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