2015 THREE(d)

Merit
Question
The equation of motion of a particle is given by the differential equation

d2xdt2=−k2x\dfrac{d^2x}{dt^2} = -k^2x

where xx is the displacement of the particle from the origin at time tt, and kk is a positive
constant.

(i) Show that x=Acos⁡kt+Bsin⁡ktx = A\cos kt + B\sin kt, where AA and BB are constants, is a solution of the
equation of motion.

(ii) The particle was initially at the origin and moving with velocity 2k2k.

Find the values of AA and BB in the solution x=Acos⁡kt+Bsin⁡ktx = A\cos kt + B\sin kt.
Official Answer
(d)(i)(d)(i)
dxdt=−Aksin⁡kt+Bkcos⁡kt\dfrac{dx}{dt}=-Ak\sin kt+Bk\cos kt

d2xdt2=−Ak2cos⁡kt−Bk2sin⁡kt\dfrac{d^2x}{dt^2}=-Ak^2\cos kt-Bk^2\sin kt
=−k2(Acos⁡kt+Bsin⁡kt)=-k^2(A\cos kt+B\sin kt)
=−k2x=-k^2x

(ii)(ii)
x(0)=0⇒Acos⁡0+Bsin⁡0=0x(0)=0\Rightarrow A\cos 0+B\sin 0=0
A=0A=0
v(0)=2k⇒2k=−Aksin⁡(0)+Bkcos⁡(0)v(0)=2k\Rightarrow 2k=-Ak\sin(0)+Bk\cos(0)
B=2B=2
Grading Criteria

Achievement (u)

  • Correct dxdt\dfrac{dx}{dt}
  • Or d2xdt2\dfrac{d^2x}{dt^2} consistent with dxdt\dfrac{dx}{dt} incorrect

Merit (r)

  • Parts (i) and (ii) both correct.

Excellence T1

-

Excellence T2

-
Exam Paper (2015)Assessment Schedule (2015)
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