InterviewSolution
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InterviewSolution
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The speed v of a particle moving along a straight line. When it is at distance x from a fixed point on the line is `v^(2)=144-9x^(2)`. Select the correct alternativesA. The magnitude of acceleration at a distance 3 units from the fixed point is 27 unitsB. the motion is simple harmonic with `T=(2pi)/(3)` unitsC. The maximum displacement from the fixed point is 4 unitsD. all are correct |
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Answer» Correct Answer - D (d) `v=omegasqrt(A^(2)+x^(2))" "`(in SHM) or `v^(2)=omega^(2)A^(2)-omega^(2)x^(2)` Comparing the given equation with this equation we get, `omega^(2)=9` `therefore omega=3=(2pi)/(T)` `therefore T=(2pi)/(3)` units Also , `omega^(2)A^(2)=144` `therefore A=4 "units", |a|=omega^(2)x=(9)(3)=27` units Displacement ` le` distance |
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