Question

# The velocity v and displacement x of a particle executing simple harmonic motion are related as vdvdx=−ω2x At x=0,v=v0. Find the velocity v when the displacement becomes x.

A

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B

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C

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D

None of these

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Solution

## The correct option is A We have, vdvdx=−ω2x or, v dv=−ω2x dx or, v∫v0v dv=x∫0−ω2x dx .......(i) When summation is made on −ω2 x dx the quantity to be varied is x. When summation is made on vdv the quantity to be varied is v. As x varies from 0 to x the velocity varies from v0 to v. Therefore, on the left the limits of integration are from v0 to v and on the right they are from 0 to x. Simplifying (i), [12v2]vv0=−ω2[x22]x0 or, 12(v2−v20)=−ω2x22 or, v2=v20−ω2x2 or, v=√v20−ω2x2

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