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What is the acceleration of the block when x= 0.160 m?

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.

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