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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.
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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Regards,
Askiitians ExpertsSoumyajit Das IIT Kharagpur
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