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The figure gives the position as function of time, of 20 g block in simple harmonic oscillation...

Question:

The figure gives the position as function of time, of 20 g block in simple harmonic oscillation on the end of a spring. The period of oscillations is 2.0 s. a. What is the amplitude of the oscillations?

b. What is the maximum speed of the oscillations?

c. What is the spring constant?

figure.1

Spring:

Spring is defined as a mechanical component that is used to store energy in the form of elastic potential energy. Different types of springs are used in a machine like helical spring, torsion spring, leaf spring, etc. The strength of the spring depends on the spring index and spring stiffness.

Answer and Explanation: 1


Given Data


  • The mass of a block is; {eq}m = 20\;{\rm{g}} {/eq}
  • The time period of oscillation is; {eq}T = 2.0\;{\rm{s}} {/eq}


(b)


From the given figure, it is clearly observed that the amplitude is {eq}A = 16\;{\rm{cm}} {/eq} that indicates the maximum displacement of the block.


Find the angular speed of the block.

{eq}\begin{align*} \omega & = \dfrac{{2\pi }}{T}\\ & = \dfrac{{2\pi }}{2\; \rm s}\\ & = {\rm{3}}{\rm{.14}}\;{\rm{rad/s}} \end{align*} {/eq}


The formula of maximum speed of oscillation of the block is,

{eq}{V_{\max }} = \omega A {/eq}


Substitute the given value in the above equation to find the maximum speed of the oscillation.

{eq}\begin{align*} {V_{\max }} &= \omega A\\ & = 3.14\;{\rm{rad/s}} \times 16\;{\rm{cm}}\left( {\dfrac{{{\rm{1}}\;{\rm{m}}}}{{{\rm{100}}\;{\rm{cm}}}}} \right)\\ & = 0.5024\;{\rm{m/s}} \end{align*} {/eq}

So, the maximum speed of the oscillation is {eq}0.5024\;{\rm{m/s}} {/eq}.


(c)


The formula of the spring constant is,

{eq}k = {\omega ^2}m {/eq}

Here, {eq}k {/eq} is the spring constant and {eq}m {/eq} is the mass of an object.


Substitute the given value in the above equation to find the spring constant.

{eq}\begin{align*} k& = {\omega ^2}m\\ &= {\left( {{\rm{3}}{\rm{.14}}\;{\rm{rad/s}}} \right)^2} \times 20\;{\rm{g}}\left( {\dfrac{{{\rm{1}}\;{\rm{kg}}}}{{{\rm{1000}}\;{\rm{g}}}}} \right)\\ &= 0.197192\;{\rm{N/m}} \end{align*} {/eq}

So, the spring constant is {eq}0.197192\;{\rm{N/m}} {/eq}.


Learn more about this topic:

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Practice Applying Spring Constant Formulas

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Chapter 17 / Lesson 11
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In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.


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