The main objective of this question is to find the **amplitude** of the **oscillating block** when t**he total energy gets doubled**.This question uses the concept of **simple harmonic motion** and the **total mechanical energy** of simple harmonic motion. The **t****otal mechanical energy** of the simple harmonic motion is equal to the **sum of total kinetic energy** and the **sum of total potential energy**.

## Expert Answer

We are **given** with:

The **amplitude of oscillating block** $= 20 \space cm$.

We have to **find the amplitude** of the **oscillating block** when the **total energy gets doubled**.

We **know** that:

\[E \space = \space K \space + \space U\]

\[\frac{1}{2}kA^2 \space = \space \frac{1}{2}mv^2 \space + \space \frac{1}{2}kx^2\]

**Mathematically,** the **total mechanical energy** is represented as:

\[E \space = \space \frac{1}{2}kA^2\]

\[E \space = \space \sqrt \frac{2E}{k} \]

**Then**:

\[A \space = \space \sqrt E\]

\[\frac{A_1}{A_2} \space = \space \frac{\sqrt E}{\sqrt 2E} \]

\[\frac{A_1}{A_2} \space = \space \frac{1}{\sqrt 2} \]

\[A_2 \space = \space \sqrt2 (20)\]

\[A_2 \space = \space 28.28 \space cm\]

## Numerical Answer

The **amplitude of the oscillating block** will be $28.28 \space cm$ when the total energy gets **doubled**.

## Example

Oscillating blocks have an amplitude of $40 \space cm$, $60 \space cm$, and $80 \space cm$. Find the amplitude of the oscillating block when the total energy gets doubled.

We are **given**:

The **amplitude of oscillating** block $= 40 \space cm$.

We have to **find** the amplitude of the **oscillating block** when the** total energy** gets **doubled**.

We **know** that:

\[E \space = \space K \space + \space U\]

\[\frac{1}{2}kA^2 \space = \space \frac{1}{2}mv^2 \space + \space \frac{1}{2}kx^2\]

**Mathematically,** the total mechanical energy is represented as:

\[E \space = \space \frac{1}{2}kA^2\]

\[E \space = \space \sqrt \frac{2E}{k} \]

**Then**:

\[A \space = \space \sqrt E\]

\[\frac{A_1}{A_2} \space = \space \frac{\sqrt E}{\sqrt 2E} \]

\[\frac{A_1}{A_2} \space = \space \frac{1}{\sqrt 2} \]

\[A_2 \space = \space \sqrt2 (40)\]

\[A_2 \space = \space 56.56 \space cm\]

Now **solving** for $60 \space cm$ amplitude.

We are **given**:

The amplitude of oscillating block $= 60 \space cm$.

We have to find the **amplitude** of the oscillating block when the **total energy** gets doubled.

We **know** that:

\[E \space = \space K \space + \space U\]

\[\frac{1}{2}kA^2 \space = \space \frac{1}{2}mv^2 \space + \space \frac{1}{2}kx^2\]

**Mathematically,** the total **mechanical energy** is represented as:

\[E \space = \space \frac{1}{2}kA^2\]

\[E \space = \space \sqrt \frac{2E}{k} \]

**Then**:

\[A \space = \space \sqrt E\]

\[\frac{A_1}{A_2} \space = \space \frac{\sqrt E}{\sqrt 2E} \]

\[\frac{A_1}{A_2} \space = \space \frac{1}{\sqrt 2} \]

\[A_2 \space = \space \sqrt2 (60)\]

\[A_2 \space = \space 84.85 \space cm\]

Now **solving** for $80 \space cm$ amplitude.

We are **given**:

The **amplitude of oscillating** block $= 80 \space cm$.

\[E \space = \space K \space + \space U\]

\[\frac{1}{2}kA^2 \space = \space \frac{1}{2}mv^2 \space + \space \frac{1}{2}kx^2\]

\[E \space = \space \frac{1}{2}kA^2\]

\[E \space = \space \sqrt \frac{2E}{k} \]

\[A \space = \space \sqrt E\]

\[\frac{A_1}{A_2} \space = \space \frac{\sqrt E}{\sqrt 2E} \]

\[\frac{A_1}{A_2} \space = \space \frac{1}{\sqrt 2} \]

\[A_2 \space = \space \sqrt2 (80)\]

\[A_2 \space = \space 113.137 \space cm\]