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A particle of mass 10g is describing SHM along a straight line with a period of 4s and amplitude 12 cm. What is the kinetic energy when it is (i) 3cm (ii) 6 cm from its equilibrium position. How do you account for the difference between the two values ? `[use pi^(2)=10]` |
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Answer» Here, `m=10g, T=4s, A=12cm.` Case (i), `y=3cm, omega=(2pi)/(T)=(2pi)/(4)=(pi)/(2)rad//s` velocity, `v=omegasqrt(A^(2)-y^(2))=(pi)/(2)sqrt(144-9)` `=(pi)/(2)sqrt(135)cm//s` `KE=(1)/(2)mv^(2)=(1)/(2)xx10xx((pi^(2))/(4)xx135)` `=168.5 erg` Case (ii), `y=6cm` `v=omegasqrt(144-36)=(pi)/(2)sqrt(108)cm//s` `KE=(1)/(2)mv^(2)=(1)/(2)xx10xx((pi^(2))/(4)xx108)=1350erg` It means the KE of particle decreases when it moves from `y=3cm ` to `6 cm.` Thi is due to increase in its potential energy. |
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