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A block attached to an ideal spring undergoes simple harmonic motion about its equilibrium position $(x=0)$ with amplitude $A .$ What fraction of the total energy is in the form of potential energy when the block is at position $x=\frac{1}{4}$ ?(A) $\frac{1}{16}$(B) $\frac{1}{4}$(C) $\frac{3}{4}$(D) $\frac{15}{16}$

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Chapter 13

Practice Test 3

Section 1

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Hi in the given problem and plead ute of oscillations off a block attached to ideal spring is given us a No. Then the displacement Of this block from the initial mean position is 1/4 of its amplitude. We have to find the fraction of its potential energy from that total energy. So it's electra's tactic. Sorry, It's elastic potential energy at this. Much displacement will be given as E. U. Is equal to half K. There K is. It's a spring factor into X square. So it becomes half-k into a by four to the whole square. Or we can say this is half K. A square by 16. Now it's total the last week. And energy means the maximum energy whether it is maximum potential energy or maximum kind of energy that is given as half. Okay into square of ham. Please do so. The fraction of potential energy will be given by E. U by E. So we can say this fraction is equal to have K. Is square by 16, divided by half K squares. For canceling this, We get to know this fraction is 1x16. Hence we can say here our option a is correct. Thank you.

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