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Question

A block undergoes simple harmonic motion about its equilibrium position ($$ x = 0$$) with amplitude $$A$$. Calculate fraction of the total energy is in the form of kinetic energy when the block is at position $$x=\dfrac{1}{2}A$$.


A
13
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B
38
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C
12
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D
23
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E
34
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Solution

The correct option is E $$\dfrac{3}{4}$$
At the endpoints of the oscillation, the displacement (x) is equal to amplitude A.  Since x is maximum so kinetic energy is zero and  the total energy  is in the form potential energy of spring i.e $$ \dfrac{1}{2}kA^2$$. 
It is also the expression of total energy at any other position. So, at $$x=\dfrac{A}{2}$$ we can write $$KE+PE=E=\dfrac{1}{2}kA^2$$
or $$KE+\dfrac{1}{2}k (\dfrac{A}{2})^2=\dfrac{1}{2}kA^2$$ or $$KE=(\dfrac{3}{8})KA^2$$
Thus, $$\dfrac{KE}{E}=\dfrac{(\dfrac{3}{8}) kA^2}{(\dfrac{1}{2})kA^2}=\dfrac{3}{4}$$


Physics

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