The period of oscillation of simple pendulum of length L suspended from the roof of the vehicle whic

Q: Why does the pendulum hang perpendicular to the incline in equilibrium?

In the vehicle's frame, the net effective gravity vector $\vec{g}_{\text{eff}}$ is perpendicular to the incline. Since the pendulum bob always aligns with the net force in equilibrium, it hangs perpendicular to the incline.

Q: Why is the angle between $\vec{g}$ and $-\vec{a}$ equal to $90^\circ + \alpha$?

The vector $\vec{g}$ points vertically downwards. The vector $-\vec{a}$ points up the incline. The angle of the incline with the horizontal is $\alpha$, which means the incline makes an angle of $90^\circ - \alpha$ with the downward vertical. Therefore, the angle between the downward vertical and the upward incline is $180^\circ - (90^\circ - \alpha) = 90^\circ + \alpha$.

Q: Can we solve this problem from the ground (inertial) frame?

Yes! In the ground frame, the bob undergoes a complex motion where it accelerates down the incline along with the vehicle while oscillating. Writing the equations of motion in the ground frame is much more tedious, which is why the accelerating frame with pseudo-forces is highly preferred.

Q: What if the inclined plane has friction?

If there is friction with coefficient $\mu$, the acceleration of the vehicle down the incline becomes $a = g(\sin\alpha - \mu\cos\alpha)$. You would substitute this new value of $a$ into the vector addition formula to find the new $g_{\text{eff}}$.

Q: Does the mass of the pendulum bob affect the time period?

No, just like a standard simple pendulum, the mass of the bob cancels out in the equations of motion, so the time period depends only on the length $L$ and the effective gravity $g_{\text{eff}}$.

Question

The period of oscillation of simple pendulum of length $L$ suspended from the roof of the vehicle which moves without friction, down an inclined plane of inclination $\alpha$, is given by

Options:

(A) $2\pi \sqrt{\frac{L}{g \cos \alpha}}$

(B) $2\pi \sqrt{\frac{L}{g \sin \alpha}}$

(C) $2\pi \sqrt{\frac{L}{g}}$

(D) $2\pi \sqrt{\frac{L}{g 	an \alpha}}$

Sigma Percentile · Accepted Answer

### Introduction to the Problem

Imagine a simple pendulum hanging from the ceiling of a stationary room. It oscillates with a familiar time period of $T = 2\pi \sqrt{\frac{L}{g}}$. But what happens when we place this entire setup inside a vehicle that is sliding down a frictionless inclined plane?

This is not just a standard pendulum problem; it is a beautiful exploration of **non-inertial reference frames** and **effective gravity**.

To solve this, we must step inside the accelerating vehicle and look at the world through its eyes. Let's embark on this thrilling physics journey!

### Analyzing the Accelerating Frame

When the vehicle slides down a frictionless inclined plane of angle $\alpha$, it is accelerated by the component of gravity acting parallel to the incline.

This acceleration is given by:

$$a = g \sin \alpha$$

Because the vehicle is accelerating, any observer inside the vehicle is in a **non-inertial reference frame**.

To apply Newton's laws or analyze simple harmonic motion from within this frame, we must introduce a **pseudo-force** on the pendulum bob of mass $m$.

This pseudo-force acts in the direction opposite to the vehicle's acceleration:

$$\vec{F}_{	ext{pseudo}} = -m\vec{a}$$

Thus, the pseudo-acceleration experienced by the bob is $-\vec{a}$, which points straight up the incline with a magnitude of $g \sin \alpha$.

### The Concept of Effective Gravity

In any accelerating frame, the pendulum bob experiences two acceleration-like effects: the real acceleration due to gravity $\vec{g}$ pointing vertically downwards, and the pseudo-acceleration $-\vec{a}$ pointing up the incline.

We can combine these two into a single vector called the **effective gravity** $\vec{g}_{	ext{eff}}$:

$$\vec{g}_{	ext{eff}} = \vec{g} - \vec{a}$$

The time period of the pendulum's small oscillations will then depend entirely on this effective gravity:

$$T = 2\pi \sqrt{\frac{L}{g_{	ext{eff}}}}$$

Our main task now is to find the magnitude of this vector sum.

### Vector Geometry and Calculation

Let's look at the geometry of the two vectors we need to add:

1. The gravity vector $\vec{g}$ acts vertically downwards.

2. The pseudo-acceleration vector $-\vec{a}$ acts up along the inclined plane.

What is the angle $	heta$ between these two vectors?

The angle of the incline with the horizontal is $\alpha$. Therefore, the angle between the downward vertical and the upward incline is exactly:

$$	heta = 90^\circ + \alpha$$

Now, we use the vector addition formula to find the magnitude of $\vec{g}_{	ext{eff}}$:

$$g_{	ext{eff}} = \sqrt{g^2 + a^2 + 2ga \cos(90^\circ + \alpha)}$$

Substitute $a = g \sin \alpha$ and $\cos(90^\circ + \alpha) = -\sin \alpha$ into the equation:

$$g_{	ext{eff}} = \sqrt{g^2 + (g \sin \alpha)^2 + 2g(g \sin \alpha)(-\sin \alpha)}$$

$$g_{	ext{eff}} = \sqrt{g^2 + g^2 \sin^2 \alpha - 2g^2 \sin^2 \alpha}$$

$$g_{	ext{eff}} = \sqrt{g^2 - g^2 \sin^2 \alpha}$$

$$g_{	ext{eff}} = \sqrt{g^2(1 - \sin^2 \alpha)}$$

Using the fundamental trigonometric identity $1 - \sin^2 \alpha = \cos^2 \alpha$, we get:

$$g_{	ext{eff}} = \sqrt{g^2 \cos^2 \alpha} = g \cos \alpha$$

This is an incredibly elegant result! The effective gravity is perpendicular to the inclined plane, and its magnitude is simply $g \cos \alpha$.

### Final Time Period

Now, we substitute this value of $g_{	ext{eff}}$ back into our time period formula:

$$T = 2\pi \sqrt{\frac{L}{g \cos \alpha}}$$

This matches **Option (a)** perfectly.

Physically, this means that the pendulum oscillates about a new equilibrium position that is perpendicular to the inclined plane, and it behaves exactly like a standard pendulum in a stationary room where gravity has been dialed down to $g \cos \alpha$.

Animated Solution for Physics - Oscillations: The period of oscillation of simple pendulum of length $L$ suspended from the roof of the vehicle which moves without friction, down an inclined plane of inclination $α$ , is given by

Select Answer:

Visualized Solution

Visualizing the Setup

The Time Period Formula in an Accelerating Frame

Finding the Acceleration of the Vehicle

Determining the Angle between $g$ and $- a$

Setting up the Vector Addition

Simplifying the Trigonometric Expression

Calculating the Final $g_{eff}$

Finding the Time Period

The Way Forward

The Sigma Insight: Force and Energy Method in SHM

Introduction to the Problem

Analyzing the Accelerating Frame

The Concept of Effective Gravity

Vector Geometry and Calculation

Final Time Period

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