Time period of a simple pendulum is T inside a lift, when the lift is stationary. If the lift moves

Q: Why do we add the acceleration $a$ to $g$ when the lift moves upwards?

When the lift accelerates upwards, the frame of reference is non-inertial. To apply Newton's laws in this frame, we must introduce a pseudo force acting in the opposite direction of the acceleration (downwards). This pseudo force ($ma$) adds to the actual weight ($mg$), making the effective gravity $g_{\text{eff}} = g + a$.

Q: What would happen if the lift was moving upwards with a constant velocity?

If the lift moves with a constant velocity, its acceleration is zero ($a = 0$). Therefore, there is no pseudo force, and the effective gravity remains $g$. The time period would not change.

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

No, the time period of a simple pendulum is independent of the mass of the bob. It only depends on the length of the pendulum and the effective acceleration due to gravity.

Q: What if the lift accelerates downwards with acceleration $g$?

If the lift accelerates downwards at $g$ (free fall), the pseudo force acts upwards with magnitude $mg$. This perfectly cancels the actual weight, making $g_{\text{eff}} = g - g = 0$. The time period would become infinite, meaning the pendulum would not oscillate.

Q: Can we use the same formula if the pendulum swings with a very large angle?

The formula $T = 2\pi\sqrt{l/g}$ is an approximation valid only for small angular displacements (typically less than 15 degrees) where $\sin\theta \approx \theta$. For larger angles, the motion is not perfectly simple harmonic, and the time period would be slightly longer.

Question

Time period of a simple pendulum is $T$ inside a lift, when the lift is stationary. If the lift moves upwards with an acceleration $g/2$, then the time period of pendulum will be

Options:

(A) $\sqrt{3}T$

(B) $\frac{T}{\sqrt{3}}$

(C) $\sqrt{\frac{3}{2}}T$

(D) $\sqrt{\frac{2}{3}}T$

Sigma Percentile · Accepted Answer

### Setting the Stage

Imagine you are standing inside an elevator, and right in front of you is a simple pendulum hanging from the ceiling. When the elevator is completely stationary, the pendulum swings smoothly back and forth. The time period of this standard swing is given by the classic formula:

$$T = 2\pi\sqrt{\frac{l}{g}}$$

Here, $l$ is the length of the string, and $g$ is the acceleration due to gravity. This is our baseline. But physics gets truly exciting when we start moving the frame of reference!

### The Concept of Effective Gravity

Now, let's hit the up button. The elevator starts accelerating upwards with an acceleration of $a = \frac{g}{2}$. Because you and the pendulum are inside an accelerating frame of reference, you will experience a **pseudo force**.

Since the elevator accelerates upwards, this pseudo force acts strictly downwards. For the pendulum bob of mass $m$, this downward pseudo force is $ma$. This force adds to the actual weight of the bob ($mg$), making it feel heavier. We capture this effect by defining a new term called **effective gravity** ($g_{	ext{eff}}$).

$$g_{	ext{eff}} = g + a$$

### The Mathematical Shift

Let's calculate exactly how much this effective gravity has increased. By substituting our given acceleration $a = \frac{g}{2}$ into the equation, we get:

$$g_{	ext{eff}} = g + \frac{g}{2} = \frac{3g}{2}$$

The effective gravity is now 1.5 times stronger! Now, we bring back our master equation for the time period, but we replace $g$ with our new $g_{	ext{eff}}$:

$$T_1 = 2\pi\sqrt{\frac{l}{g_{	ext{eff}}}}$$

Substituting the value we just found:

$$T_1 = 2\pi\sqrt{\frac{l}{\frac{3g}{2}}} = 2\pi\sqrt{\frac{2l}{3g}}$$

### The Final Verdict

To see how this new time period $T_1$ compares to our original time period $T$, we need to separate the constants. Let's pull the fraction $\sqrt{\frac{2}{3}}$ out of the square root:

$$T_1 = \sqrt{\frac{2}{3}} \left( 2\pi\sqrt{\frac{l}{g}} ight)$$

Notice that the term inside the parentheses is exactly our original time period $T$. Therefore, we arrive at our final, elegant relationship:

$$T_1 = \sqrt{\frac{2}{3}} T$$

The time period has decreased, meaning the pendulum swings faster because it feels a stronger downward pull. Always remember, the behavior of a pendulum is intimately tied to the acceleration of its environment!

Animated Solution for Physics - Oscillations: Time period of a simple pendulum is $T$ inside a lift, when the lift is stationary. If the lift moves upwards with an acceleration $g /2$ , then the time period of pendulum will be

Select Answer:

Visualized Solution

$T = 2 π \frac{l}{g}$

$T = 2 π \frac{l}{g _{eff}}$

$a = \frac{g}{2} ↑$

$g_{eff} = g + a$

$T_{1} = 2 π \frac{l}{g _{eff}}$

$T_{1} = \frac{2}{3} T$

\text{What if } a = g \downarrow?

The Sigma Insight: Simple Harmonic Motion (SHM)

Setting the Stage

The Concept of Effective Gravity

The Mathematical Shift

The Final Verdict

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