A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length o

Q: Why is the kinetic energy maximum at the mean position?

In Simple Harmonic Motion, the restoring force pulls the object towards the mean position. As it approaches the mean position, it accelerates, reaching its maximum velocity exactly at the center. Since kinetic energy depends on the square of velocity ($K = \frac{1}{2}mv^2$), it is also maximum at this point.

Q: Does the mass of the pendulum bob affect the final ratio?

No. While the absolute value of the kinetic energy does depend on the mass, the problem asks for a ratio between two states of the pendulum. Assuming the same bob is used (or bobs of equal mass), the mass $m$ acts as a constant and completely cancels out when you take the ratio $K_1 / K_2$.

Q: What is the difference between linear amplitude and angular amplitude?

Linear amplitude ($a$) is the actual physical distance measured along the arc of the swing (e.g., 5 cm). Angular amplitude ($\theta_0$) is the maximum angle the string makes with the vertical (e.g., 10 degrees). They are related by the arc length formula: $a = l\theta_0$.

Q: Why did we use $\omega = \sqrt{g/l}$ instead of $\omega = \sqrt{k/m}$?

$\omega = \sqrt{k/m}$ is the general angular frequency for a spring-mass system. For a simple pendulum, the restoring force is provided by gravity, and the effective 'spring constant' is $k = mg/l$. Substituting this into $\sqrt{k/m}$ gives the specific pendulum formula $\omega = \sqrt{g/l}$.

Q: If the length is doubled, what happens to the time period of the pendulum?

The time period of a simple pendulum is given by $T = 2\pi\sqrt{l/g}$. Since $T$ is directly proportional to the square root of the length ($\sqrt{l}$), doubling the length will increase the time period by a factor of $\sqrt{2}$ (approximately 1.414 times longer).

Question

A pendulum is executing simple harmonic motion and its maximum kinetic energy is $K_1$. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is $K_2$. Then

Options:

(A) $K_2 = 2K_1$

(B) $K_2 = \frac{K_1}{2}$

(C) $K_2 = \frac{K_1}{4}$

(D) $K_2 = K_1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine two simple pendulums swinging side by side. The first pendulum has a length $l_1$, and the second pendulum is exactly twice as long, meaning $l_2 = 2l_1$. The problem explicitly states that both pendulums are swinging with the exact same linear amplitude, $a$. Our goal is to figure out how the maximum kinetic energy of the second pendulum, $K_2$, compares to the maximum kinetic energy of the first pendulum, $K_1$.

### The Master Equation
To solve this, we need to recall the fundamental energy equation for a particle executing Simple Harmonic Motion (SHM). The kinetic energy reaches its absolute maximum when the pendulum passes through its lowest point (the mean position). The formula for this maximum kinetic energy is:

$$K_{max} = \frac{1}{2} m \omega^2 a^2$$

Here, $m$ is the mass of the pendulum bob, $\omega$ is the angular frequency of the oscillation, and $a$ is the linear amplitude.

Now, we know that the angular frequency $\omega$ for a simple pendulum depends purely on gravity and its length. The relationship is given by:

$$\omega = \sqrt{\frac{g}{l}}$$

If we square both sides, we get $\omega^2 = \frac{g}{l}$. Let's substitute this expression for $\omega^2$ back into our kinetic energy equation. This gives us a much more useful form of the equation tailored specifically for a simple pendulum:

$$K_{max} = \frac{1}{2} m \left(\frac{g}{l}ight) a^2$$

### Establishing Proportionality
Let's look closely at this new expression. The problem tells us that the amplitude $a$ is the same for both pendulums. The mass $m$ and the acceleration due to gravity $g$ are also constants. The only variable that is changing between the two pendulums is the length $l$, which is sitting right there in the denominator.

This tells us a crucial physical truth for this specific scenario: **The maximum kinetic energy is inversely proportional to the length of the pendulum.**

$$K_{max} \propto \frac{1}{l}$$

### Final Calculation
Because of this inverse proportionality, we can easily set up a ratio comparing the two kinetic energies. The ratio of $K_1$ to $K_2$ will be equal to the inverse ratio of their lengths:

$$\frac{K_1}{K_2} = \frac{l_2}{l_1}$$

We are given that the second pendulum is twice as long as the first, so we substitute $l_2 = 2l_1$ into our ratio:

$$\frac{K_1}{K_2} = \frac{2l_1}{l_1} = 2$$

Finally, rearranging this equation to solve for $K_2$, we get our answer:

**$$K_2 = \frac{K_1}{2}$$**

The longer pendulum, despite swinging the same linear distance, is moving slower at its lowest point, resulting in exactly half the maximum kinetic energy!

### A Word of Caution
Always read the wording carefully! If the question had stated that the **angular amplitude** ($	heta_0$) was constant instead of the linear amplitude ($a$), the outcome would be entirely different. Since linear amplitude $a = l	heta_0$, substituting this into the energy equation would make $K_{max} \propto l$. In that alternate reality, the longer pendulum would have *more* kinetic energy. Pay attention to the details!

Select Answer:

Visualized Solution

Visualizing the Pendulums

Formula for Maximum Kinetic Energy

Angular Frequency of a Pendulum

Substituting $ω^{2}$ into $K_{ma x}$

Establishing Proportionality

Setting up the Ratio

Final Calculation

The Catch: Linear vs Angular Amplitude

The Sigma Insight: Simple Harmonic Motion

Analyzing the Setup Imagine two simple pendulums swinging side by side

The Master Equation To solve this, we need to recall the fundamental energy equation for a particle executing Simple Harmonic Motion (SHM)

Establishing Proportionality Let's look closely at this new expression

Final Calculation Because of this inverse proportionality, we can easily set up a ratio comparing the two kinetic energies

A Word of Caution Always read the wording carefully! If the question had stated that the angular amplitude ( $θ_{0}$ ) was constant instead of the linear amplitude ( $a$ ), the outcome would be entirely different

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