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The total energy ofa particle of mass 0.1 kg performing SHM is 0.2 J . Find its maximum speed and period if the amplitude is 2cm. |
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Answer» Solution :Data : m= 0.1 kg, E = 0.2 J , A = 2 cm = ` 2XX 10^(-2)` m (i) The TOTAL ENERGY of the particle. ` E = KE_(max) = 1/2mv^(2) _(max)` `v_(max) = sqrt((2E)/m) = sqrt((2xx0.2)/(0.1)= 2 m//s` (ii) ` omega = (2PI)/T = omegaA = v_(max) = 2 m//s` ` (2piA)/T= 2m//s` The period of SHM of the particle . ` T= (2piA)/2 = piA = 3.142 xx 2xx 10^(-2) = 0.06284 s` |
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