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a particle of mass 0.50 kg executes a simple harmonic motion under a for F = -(50N/m)x. if it crosses the centre of oscillation with a speed of 10m/s find the amplitude of the motion.

sudeep joseph , 15 Years ago
Grade 11
anser 1 Answers
Askiitians Expert Soumyajit IIT-Kharagpur

Last Activity: 15 Years ago

Dear Sudeep Joseph,

Ans:- The energy at the mean position is totally kinetic and at max amplitude it is totally potential now the potential energy may be calculated like this work done in moving a small distance dx against the forceis  = -F dx=50 x dx leet A be the amplitude then integrating the function within 0 to A we get, 1/2(50 A²) is the PE now it should be equal to the KE as the total energy is conserved in SHM. So, solving we get (0.5)(10)²=50 A²

or A=1 meter(ans)

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Soumyajit Das IIT Kharagpur

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