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A spring of force constant `1200Nm^(-1)` is mounted on a horizontal table as shown in figure. A mass of 3.0kg is attached to the free end of the spring, pulled side ways to a distance of 2.0cm and released , what is (a) the speed of the mass when the spring is compressed by 1.0cm? Potential energy of the mass when it momentarily comed to rest ? (c) Total energy of the oscillating mass. |
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Answer» `m=3kg,k=1200Nm^(-1)`, `a=0.92m` (a) `V+?, y=1.0cm=0.01m. ` Let , `y=asinomegat` `=aomegasqrt(1-sin^(2)omegat)` or `V=aomega(1-y^(2)//a^(2))^(1//2)` `=omegasqrt(a^(2)-y^(2))=sqrt((k)/(m))sqrt((a^(2)-y^(2)))` `=sqrt((1200)/(3))[(0.02)^(2)-(0.01)^(2)]^(1/(2))` (b) Max. P.E., `U_(max)-(1)/(2)momega^(2)y^(2)_(max)` `=(1)/(2)momega^(2)a^(2)=(1)/(2)ka^(2)=(1)/(2)xx1200xx(0.02)^(2)` `=0.24J` (c) At the extreme position, where `y=a` the mass just comes to res, its K.E. becomes zero `:.` Total energy `=P.E. +K.E. =0.24+0` `=0.24J` |
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