Question

5. A particle of mass m is subject to the force F-asin(bx). (a) If the maximum value of the corresponding potential energy is g, what are the turning points for a particle of energy E ,? (b) Find the speed of the particle as a function of position, if the paosition of the partide as a function of time

Answer 1

We know F(x) = -d(U)/dx, F = a sin(bx) U = integral dU = integral -F(x)dx = integral -a sin(bx)dx = a/b cos(bx) + c As given max value of U is a/b and should be when cos(bx) = 1 U = a/b times 1 + c = a/b + 0 Therefore U = a/b cos(bx) equation (1) (a) graph of U vs x, given E = a/2b (b) particle starts at rest at one tuning pt E = a/2b = K + U = 0 + U at tuning pt E = a/2b = 1/2 mv^2 + a/b cos(bx) = K + U a/b (1/2 - cos(bx)) = 1/2 mv^2 a/b (1 - 2 cos(bx)) = mv^2 \r\n v = squareroot a/bm (1 - 2 cos(bx)) m/s Ans (c) a_acc = F/m = a/m sin(bx) v = integral a_acc dt = integral a/m sin(bx)dt = a/m sin(bx)t + c_1 [as x const wrt t] where