A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of th

Q: Why do we differentiate the formula instead of just plugging in the values?

Differentiation is a powerful tool for finding the relationship between very small changes (perturbations) in variables without having to do complex binomial expansions or square root approximations. It gives us a direct linear relationship.

Q: Why is $\Delta g = 2 \omega_s^2 A$ and not just $\omega_s^2 A$?

The support oscillates up and down. At the top extreme, the acceleration is downwards, making $g_{eff} = g - \omega_s^2 A$. At the bottom extreme, the acceleration is upwards, making $g_{eff} = g + \omega_s^2 A$. The total peak-to-peak variation is the difference between these two, which is $2 \omega_s^2 A$.

Q: Why did we substitute $g = \omega^2 l$ in the final steps?

The problem does not explicitly give us the value of $g$. While we could infer that $g = 100 \text{ m/s}^2$ from the given $\omega = 10$ and $l = 1$, substituting $g = \omega^2 l$ algebraically eliminates $g$ from the equation, making the final calculation cleaner and less prone to assumption errors.

Q: The question asks for 'relative change', shouldn't the answer be dimensionless?

Yes, strictly speaking, 'relative change' means $\frac{\Delta \omega}{\omega}$, which has no units. However, this is a classic case of poor phrasing by the examiners. The options all have units of $\text{rad/s}$, which tells us we must calculate the absolute change $\Delta \omega$.

Q: What would happen if the support oscillated horizontally instead of vertically?

If the support oscillated horizontally, the effective gravity would be $g_{eff} = \sqrt{g^2 + a^2}$. For small accelerations, this change is a second-order effect (proportional to $a^2$), whereas vertical oscillation is a first-order effect (proportional to $a$).

Question

A simple pendulum of length $1 	ext{ m}$ is oscillating with an angular frequency $10 	ext{ rad/s}$. The support of the pendulum starts oscillating up and down with a small angular frequency of $1 	ext{ rad/s}$ and an amplitude of $10^{-2} 	ext{ m}$. The relative change in the angular frequency of the pendulum is best given by

Options:

(A) $1 	ext{ rad/s}$

(B) $10^{-5} 	ext{ rad/s}$

(C) $10^{-3} 	ext{ rad/s}$

(D) $10^{-1} 	ext{ rad/s}$

Sigma Percentile · Accepted Answer

### The Elevator Analogy
Imagine you are standing in an elevator. When the elevator accelerates upwards, you feel heavier. When it accelerates downwards, you feel lighter. This sensation is due to the pseudo force acting on your body, which effectively changes the gravity you experience.

Exactly the same phenomenon occurs with our simple pendulum in this problem. The support of the pendulum is oscillating vertically, which means it is constantly accelerating up and down. This vertical acceleration modifies the effective gravity $g_{eff}$ acting on the pendulum bob.

### The Calculus of Small Changes
We know that the angular frequency of a simple pendulum is given by the formula:
$$ \omega = \sqrt{\frac{g_{eff}}{l}} $$

Since the length $l$ is constant, any small change in the effective gravity $g_{eff}$ will induce a small change in the angular frequency $\omega$. To find the relationship between these small changes, we can use differentiation. Differentiating $\omega$ with respect to $g$, we get:
$$ \frac{d\omega}{dg} = \frac{1}{2\sqrt{g l}} $$

Rearranging this and dividing by the original $\omega$, we obtain a beautiful and highly useful relation for error analysis and small perturbations:
$$ \frac{d\omega}{\omega} = \frac{dg}{2g} $$

This tells us that the fractional change in the angular frequency is exactly half of the fractional change in the effective gravity.

### Peak-to-Peak Variation
The support is executing Simple Harmonic Motion (SHM) with an angular frequency $\omega_s$ and an amplitude $A$. The maximum acceleration of this support is $a_{max} = \omega_s^2 A$.

Because the support oscillates up and down, the effective gravity swings from a minimum of $(g - a_{max})$ to a maximum of $(g + a_{max})$. Therefore, the total peak-to-peak change in the effective gravity is:
$$ \Delta g = 2 a_{max} = 2 \omega_s^2 A $$

### The Final Calculation
Now, we substitute our peak-to-peak $\Delta g$ back into our fractional change equation:
$$ \frac{\Delta \omega}{\omega} = \frac{2 \omega_s^2 A}{2g} = \frac{\omega_s^2 A}{g} $$

To find the absolute change $\Delta \omega$, we multiply both sides by $\omega$:
$$ \Delta \omega = \frac{\omega \cdot \omega_s^2 A}{g} $$

We also know that for the pendulum, $g = \omega^2 l$. Substituting this into our equation allows us to eliminate $g$ entirely:
$$ \Delta \omega = \frac{\omega \cdot \omega_s^2 A}{\omega^2 l} = \frac{\omega_s^2 A}{\omega l} $$

Finally, we plug in the given values: $\omega_s = 1 	ext{ rad/s}$, $A = 10^{-2} 	ext{ m}$, $\omega = 10 	ext{ rad/s}$, and $l = 1 	ext{ m}$:
$$ \Delta \omega = \frac{(1)^2 	imes 10^{-2}}{10 	imes 1} = 10^{-3} 	ext{ rad/s} $$

**A Word on Terminology:** The question asks for the "relative change", which strictly speaking should be the dimensionless ratio $\frac{\Delta \omega}{\omega}$. However, the options are given in units of $	ext{rad/s}$, which is a massive hint that the examiner actually wants the absolute peak-to-peak change $\Delta \omega$. Always let the units guide your final interpretation!

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Visualized Solution

$g_{e f f} in an Oscillating Frame$

$Angular Frequency of Pendulum$

$Differentiating ω$

$Peak-to-Peak Change in g_{e f f}$

$Total Change in ω$

$Calculating Δ ω$

$Reflections$

The Sigma Insight: Simple Harmonic Motion (SHM)

The Elevator Analogy Imagine you are standing in an elevator

The Calculus of Small Changes

Peak-to-Peak Variation The support is executing Simple Harmonic Motion (SHM) with an angular frequency $ω_{s}$ and an amplitude $A$

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } geff​ in an Oscillating Frame

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Frequency of Pendulum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Differentiating ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Peak-to-Peak Change in geff​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Change in ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Δω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reflections