2012 Semifinal Exam Part B 13 Part B Question B1 A particle of mass m moves under a force similar to that of an ideal spring,except that the force repels the particle from the origin: F =+ma2x In simple harmonic motion,the position of the particle as a function of time can be written x(t)=A coswt+B sinwt Likewise,in the present case we have x(t)=A f(t)+B f2(t) for some appropriate functions fi and f2. a.fi(t)and f2(t)can be chosen to have the form e".What are the two appropriate values of r? b.Suppose that the particle begins at position x(0)=zo and with velocity v(0)=0.What is x(t)? c.A second,identical particle begins at position x(0)=0 with velocity v(0)=vo.The second particle becomes closer and closer to the first particle as time goes on.What is vo? Solution a.We have ma=ma2x d2x dt2 -a2x=0 As with the case of simple harmonic motion,we insert a trial function,in this case x(t)=Aert: Ae"t-a2Aert=0 d r2Aert-a2Aert=0 r2-a2=0 r=土a b.We have I(t)=Aeat+Be-at and therefore v(t)=aAeat-aBe-at Inserting our initial values, x(0)=A+B=x0 Copyright C2012 American Association of Physics Teachers2012 Semifinal Exam Part B 13 Part B Question B1 A particle of mass m moves under a force similar to that of an ideal spring, except that the force repels the particle from the origin: F = +mα2x In simple harmonic motion, the position of the particle as a function of time can be written x(t) = A cos ωt + B sin ωt Likewise, in the present case we have x(t) = A f1(t) + B f2(t) for some appropriate functions f1 and f2. a. f1(t) and f2(t) can be chosen to have the form e rt. What are the two appropriate values of r? b. Suppose that the particle begins at position x(0) = x0 and with velocity v(0) = 0. What is x(t)? c. A second, identical particle begins at position x(0) = 0 with velocity v(0) = v0. The second particle becomes closer and closer to the first particle as time goes on. What is v0? Solution a. We have ma = mα2x d 2x dt2 − α 2x = 0 As with the case of simple harmonic motion, we insert a trial function, in this case x(t) = Aert: d 2 dt2 Aert − α 2Aert = 0 r 2Aert − α 2Aert = 0 r 2 − α 2 = 0 r = ±α b. We have x(t) = Aeαt + Be−αt and therefore v(t) = αAeαt − αBe−αt Inserting our initial values, x(0) = A + B = x0 Copyright c 2012 American Association of Physics Teachers