This question aims to find the **acceleration** of the **block** attached to a **spring** that is moving along a **frictionless horizontal surface.**

This block follows the simple harmonic motion along the horizontal direction. **Simple harmonic motion** is the type of **“to and fro”** motion in which the object displaced from its mean position by an **acting force** comes back to its mean position after it covers a certain **distance**.

The **mean position** in simple harmonic motion is the **starting position** while the **extreme position** is the position in which an object covers its **maximum displacement**. When that object reaches its maximum displacement, it comes back to its starting point and this motion repeats itself.

## Expert Answer

We have to find the acceleration of the moving block on the horizontal frictionless surface. The amplitude and time of this simple harmonic motion are given.

\[ Amplitude = 0. 240 \]

\[ Time taken = 3. 08 s \]

The **position** of the block on the horizontal frictionless surface is given by **x**:

\[ x = 0. 160 m \]

We will find the **Acceleration of the block** from the angular frequency that is given by the formula:

\[ \omega = \frac { 2 \pi } { T } \]

\[ \alpha = – \omega ^ 2 x \]

By putting angular frequency in the acceleration formula. **Angular frequency** is defined as the frequency of the object in an angular motion per unit time.

\[ \alpha = – ( \frac { 2 \pi } { T } ) ^ 2 x \]

By putting the values of **time** and **position** of the block to find acceleration:

\[ \alpha = – ( \frac { 2 \pi } { 3. 08 s } ) ^ 2 ( 0. 160 m) \]

\[ \alpha = – ( 2. 039 ra \frac { d } {s} ) ^ 2 ( 0. 160 m) \]

\[ \alpha = 0. 665 \frac { m } { s ^ 2 } \]

## Numerical Results

**The acceleration of the block attached to a spring that is moving on the frictionless horizontal surface is $ 0. 665 \frac { m } { s ^ 2 } $.**

## Example

Find the** acceleration** of the **same block** when it is placed at the **position** of **0.234 m.**

The position of the block on the horizontal frictionless surface is given by x:

\[ x = 0.234 m \]

\[ \omega = \frac { 2 \pi } { T } \]

\[ \alpha = – \omega ^ 2 x \]

By putting angular frequency in the acceleration formula:

\[ \alpha = – ( \frac { 2 \pi } { T } ) ^ 2 x \]

By putting the values of time and position of the block to find acceleration:

\[ \alpha = -( \frac { 2 \pi } { 3. 08 s } ) ^ 2 ( 0.234 m) \]

\[ \alpha = -( 2. 039 ra \frac { d } {s} ) ^ 2 ( 0.234 m) \]

\[ \alpha = 0. 972 \frac { m } { s ^ 2 } \]

*Image/Mathematical drawings are created in Geogebra**.*