A simple pendulum is set up in a trolley which moves to the right with an acceleration a on a horizontal plane. Then the thread of the pendulum in the mean position makes an angle θ with the vertical :

Question

A simple pendulum is set up in a trolley which moves to the right with an acceleration a on a horizontal plane. Then the thread of the pendulum in the mean position makes an angle θ with the vertical : (A) tan- 1(a/g) in the forward direction (B) tan- 1(a/g) in the backward direction (C) tan- 1(g/a) in the backward direction (D) tan- 1(g/a) in the forward direction

Tardigrade Admin · Accepted Answer

In accelerated frame of reference, a fictitious
force (pseudo force)

ma

acts on the bob of pendulum as shown in figure.
Solution

Hence,

t an θ = \frac{ma}{m g} = \frac{a}{g} \Rightarrow θ = (t an)^{- 1} (\frac{a}{g})

in the
backward direction.

Q. A simple pendulum is set up in a trolley which moves to the right with an acceleration $a$ on $a$ horizontal plane. Then the thread of the pendulum in the mean position makes an angle $θ$ with the vertical :

$t a n^{- 1} \frac{a}{g}$ in the forward direction

$t a n^{- 1} \frac{a}{g}$ in the backward direction

$t a n^{- 1} \frac{g}{a}$ in the backward direction

$t a n^{- 1} \frac{g}{a}$ in the forward direction

In accelerated frame of reference, a fictitious
force (pseudo force) $ma$ acts on the bob of pendulum as shown in figure.

Hence, $t an θ = \frac{ma}{m g} = \frac{a}{g} \Rightarrow θ = (t an)^{- 1} (\frac{a}{g})$ in the
backward direction.

Q. A simple pendulum is set up in a trolley which moves to the right with an acceleration a on a horizontal plane. Then the thread of the pendulum in the mean position makes an angle θ with the vertical :

tan−1ga​ in the forward direction

tan−1ga​ in the backward direction

tan−1ag​ in the backward direction

tan−1ag​ in the forward direction