# A block of mass 400 \ gm is hung vertically with the help of a massless spring. Initially block...

## Question:

A block of mass {eq}400 \ g {/eq} is hung vertically with the help of a massless spring. Initially block is in equilibrium position. Spring constant of the spring is {eq}k = 200 \ N/m {/eq}. Natural length of the spring is {eq}40 \ cm {/eq}. Now the block is stretched to some distance such that the total length of the spring becomes {eq}45 \ cm {/eq} and then released. Determine the amplitude of simple harmonic motion of the block.

## Spring:

Spring is a mechanical device that uses to store energy. Whenever we applied a load on the spring, the spring gets compressed and store energy. It will release the energy and will come in its natural form to do work.

## Answer and Explanation: 1

**Given data**

- The mass of the block is: {eq}m = 400\;{\rm{g}} = 0.4\;{\rm{kg}} {/eq}

- The spring constant is: {eq}k = 200\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}} {/eq}

- The initial spring length is: {eq}{x_1} = 40\;{\rm{cm}} = 0.4\;{\rm{m}} {/eq}

- The final spring length is: {eq}{x_2} = 45\;{\rm{cm}} = 0.45\;{\rm{m}} {/eq}

The total stretch in the spring is as follows,

{eq}\begin{align*} x &= {x_2} - {x_1}\\ x &= 0.45\;{\rm{m}} - 0.4\;{\rm{m}}\\ x &= 0.05\;{\rm{m}} \end{align*} {/eq}

The natural frequency of the spring mass system is as follows,

{eq}\omega = \sqrt {\dfrac{k}{m}} {/eq}

Substitute all the values in the above equation.

{eq}\begin{align*} \omega &= \sqrt {\dfrac{{200\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}}}}{{0.4\;{\rm{kg}}}}} \\ \omega &= 22.36\;{{{\rm{rad}}} {\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right. } {\rm{s}}} \end{align*} {/eq}

The velocity of the block from the energy system of the spring is as follows,

{eq}\dfrac{1}{2}m{v^2} = \dfrac{1}{2}k{x^2} {/eq}

Substitute all the values in the above equation.

{eq}\begin{align*} \dfrac{1}{2} \times 0.4\;{\rm{kg}} \times {v^2} &= \dfrac{1}{2} \times 200\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}} \times {\left( {0.05\;{\rm{m}}} \right)^2}\\ v &= 1.118\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} \end{align*} {/eq}

The amplitude of the simple harmonic motion of the spring is as follows,

{eq}{v^2} = {\omega ^2}\left( {{A^2} - {x^2}} \right) {/eq}

Substitute all the values in the above equation.

{eq}\begin{align*} {\left( {1.118\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}} \right)^2} &= {\left( {22.36\;{{{\rm{rad}}} {\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right. } {\rm{s}}}} \right)^2}\left[ {{A^2} - {{\left( {0.05\;{\rm{m}}} \right)}^2}} \right]\\ A &= 0.0707\;{\rm{m}}\\ A &= 7.07\;{\rm{cm}} \end{align*} {/eq}

Thus, the amplitude of the spring in the harmonic motion is {eq}7.07\;{\rm{cm}} {/eq}.

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Chapter 4 / Lesson 19After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.