Q.A simple pendulum of length $l$ and having a bob of mass $M$ is suspended in a car. The car is moving on a circular track of radius $R$ with a uniform speed $v$. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

The key idea is that the effective gravity inside the car is the vector sum of actual gravity and the centrifugal acceleration due to the circular motion.

Reasoning:

1. The car moves on a circular track of radius $R$ at speed $v$. For an observer in the car, a centrifugal force $M v^2 / R$ acts radially outward on the bob.
2. When the pendulum oscillates in the radial direction, the effective acceleration $g_{\text{eff}}$ is the resultant of $g$ (downward) and the centrifugal acceleration $v^2 / R$ (horizontal, outward).
3. The magnitude of this effective acceleration is: …

The key insight here is that the pendulum is not in an inertial frame. The car is moving on a circular track, so from the car’s perspective (the non-inertial frame where the pendulum is observed), there is a centrifugal force acting on the bob. This force modifies the effective gravity that determines the pendulum’s restoring torque.

When the pendulum oscillates in the radial direction (i.e., along the line joining the car to the centre of the track), the centrifugal acceleration acts horizontally outward, perpendicular to the actual gravity. The bob therefore experiences a resultant acceleration that is the vector sum of $g$ downward and $v^2/R$ outward.

Let’s work through the steps.

1. Identify the forces in the car’s frame.

   In the car’s frame, the bob feels:

   • Its weight $Mg$ downward.
   • A centrifugal force $M v^2 / R$ outward (radially away from the track’s centre). There is no Coriolis force because the pendulum oscillates radially and the car’s rotation is uniform — but even if there were, for small oscillations the Coriolis effect is negligible in the radial direction.

2. Find the effective gravity.

   The two accelerations are perpendicular: $g$ vertical, $v^2/R$ horizontal. The magnitude of the effective gravitational acceleration $g_\text{eff}$ is:

The direction of $g_\text{eff}$ is at an angle $\theta_0$ from the vertical, where $\tan\theta_0 = \frac{v^2}{Rg}$. This is the equilibrium position of the pendulum — it hangs tilted outward.

3. Small oscillations about the new equilibrium.

   For small angular displacements from this tilted equilibrium, the restoring torque is provided by the component of $g_\text{eff}$ perpendicular to the string. The situation is exactly like a simple pendulum in a uniform effective gravity $g_\text{eff}$. The length remains $l$, and the mass cancels out.

4. Write the time period. …

Step 1: In the car's rotating (non-inertial) frame, the bob experiences its weight $Mg$ vertically downward plus a centrifugal pseudo-force $Mv^2/R$ horizontally outward.

Step 2: These two accelerations, $g$ and $v^2/R$, are mutually perpendicular, so the effective gravitational field the bob actually feels has magnitude $g_{\text{eff}}=\sqrt{g^2+(v^2/R)^2}$. …