A bob of mass `m` is suspended from the ceiling of a train moving with an acceleration a as shown in figure. Find the angle `theta` in equilibrium position.

Options

A. `theta= tan^(-1) ((g)/(a))`
B. `theta= tan^(-1) ((-2a)/(g))`
C. `theta= sin^(-1) ((a)/(g))`
D. `theta= tan^(-1) ((a)/(g))`

Answer

D

Explanation

This problem can also be solved by both the methods.
(i)Inertia Frame of Reference (Ground) FBD of bob w.r.t. ground ( only real forces) bob is also moving with an acceleration a

`therefore sumF_(x) = 0 implies T sin theta=ma`.........(i)
and `sum F_(y)=0 imples T cos theta = mg`...........(ii)
From Eqs. (i) and (ii), we get
`tan theta=(a)/(g) " or " theta =tan ^(-1)((a)/(g))`
(ii)Non-inertial Frame of Reference (Train) FBD of bob w.r.t. train (real forces+ pseudo forces)

with respect to train, bob is in equilibrium
`therefore sumF_(x)=0 implies T sin theta=mg `....(iii)
`therefore sumF_(y)= 0 implies T cos theta=mg`....(iv)
From Eqs. (iii) and (iv), we get the same result , i.e.
`theta= tan^(-1) ((a)/(g))`