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the potential energy of theparticle moving along the x-axis is given by U(X)=8x^2+2x^4 where U is in the joule and x is in m.if total mechinal energy is 9j,then limits of motion are
6 years ago
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Hi, the given equation is for a conservative force only, we can get the equation of force wrt x but its not important here. Now since potential energy is defined only for conservative forces and total mechanical energy is conserved, hence the sum of Kinetic and Potential energy is constant = 9J. Also, U(x) = 8x^2 + 2x^4. Diffrentiating, dU/dx = 16x + 8x^3. For finding maxima and minima, the diff. should be 0. hence 8x(2+x^2) = 0 which gives x= 0 as the only solution, which is the minimum value of potential energy (=0, obviously). Here Kinetic energy is max. Also note that there are no other points obtained from this differentiation, so no maxima is defined else where, except where potential energy is max. and Kinetic energy becomes 0. We get the required vaues from the initial eqn. =>8x^2 + 2x^4 = 9 (which is the max value, that either KE or PE can take.) taking x^2 as t, we get a quadratic equation, which on solving gives: t = x^2 = (-8+/- 2v(34))/4 (check it -b +/- v(b^2-4ac)/2a). Here we neglect the -ve sign becausex^2 can't be -ve. Hence we take only the +ve value of this Q solution. Now this is the value of x^2, take roots of this value again and you get the values of the limits, both +ve and -ve value of x are to be taken. (limits are when U is max for +ve x and for -ve x, for both conditions KE is 0). Here finally on solving x^2, you get : x = +/- 0.95 m(approx). Thanks
Dear student Mr. Jauneet; The given equation is for a conservative force only, we can get the equation of force wrt x but its not important here. Now since potential energy is defined only for conservative forces and total mechanical energy is conserved, hence the sum of Kinetic and Potential energy is constant = 9J. Also, U(x) = 8x^2 + 2x^4. Diffrentiating, dU/dx = 16x + 8x^3. For finding maxima and minima, the diff. should be 0. hence 8x(2+x^2) = 0 which gives x= 0 as the only solution, which is the minimum value of potential energy (=0, obviously). Here Kinetic energy is max. Also note that there are no other points obtained from this differentiation, so no maxima is defined else where, except where potential energy is max. and Kinetic energy becomes 0. We get the required vaues from the initial eqn. =>8x^2 + 2x^4 = 9 (which is the max value, that either KE or PE can take.) taking x^2 as t, we get a quadratic equation, which on solving gives: t = x^2 = (-8+/- 2v(34))/4 (check it -b +/- v(b^2-4ac)/2a). Here we neglect the -ve sign becausex^2 can't be -ve. Hence we take only the +ve value of this Q solution. Now this is the value of x^2, take roots of this value again and you get the values of the limits, both +ve and -ve value of x are to be taken. (limits are when U is max for +ve x and for -ve x, for both conditions KE is 0). Here finally on solving x^2, you get : x = +/- 0.95 m(approx). Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards, Askiitians Experts Pramod Kumar IITR Alumni
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