1mg(h+x)= 1/2 kx^2 , x=amplitude. energy= 1/2 k x^2, since energy remains unchanged (considering lowest pt. having 0 P.E.)
A body of mass m fell from a height h onto the pan of a spring balance. The masses of the pan and the spring are negligible, the stiffness of the latter is k. Having stuck to the pan, the body starts performing harmonic oscillations in the vertical direction. Find the amplitude and the energy of the oscillations
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8 Answers
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62will x be the amplitude?
I presume that u have taken x to be the compression in the spring. from the initial position
1sorry, i actuall thought a balance hanging from spring...
i think if its compressing the spring, then...
amplitude= x- y , where mg=ky
62yes now u are right..
sorry i did not post the image.. (this was another problem from irodov.. i gave it only for this particular mistake that u made!) But my wrong.. i should have put the image.
1let v=√2gh
when it gets stuck , extension of the spring =0 and taking it as the reference point potential energy 0
let it stop at a distance x downwards
then = mv2/2=kx2/2-mgx
we get x=2mg+√(4m2g2+4mv2k)/2k
distance of point where force =0
mg=ky y=mg/k
x-y = amplitude = (1/k)√(m2g2+mv2k)
energy = mv2/2(as at that moment of time potential energy= 0 and compression = 0 )
in all the above cases v=√2gh
