Question 14:
--- 14(a) ---
14(a) | Models the motion of the particle
by forming a second order
differential equation. (must have
correct terms but allow sign errors | AO3.3 | M1 | d2x dx
M = 4M − 8Mx −
dt2 dt
d2x dx
+4 +8x =0
dt2 dt
λ2 +4λ+8=0
λ=−2±2i
Complex roots General solution
is of the form:
x= Ae−2tcos(2t ⇒+B)
x=−2Ae−2tcos(2t+B)−2Ae−2tsin(2t+B)
x(0)=0
so,
−2Acos (B)−2Asin (B)=0
⇒tan B=−1
π
⇒ B=−
4
x(0)=1
A
⇒ =1
2
⇒ A= 2
x= 2e−2tcos(2t− π )
4
∴
the particle oscillates about O,
with period π seconds and
∴decreasing amplitude.
Obtains correct differential
equation | AO1.1b | A1
Forms and solves auxiliary
equation for ‘their’ D.E. | AO1.1a | M1
States a correct form of the
general solution for ‘their’ auxiliary
solution. (ft only if both M1 marks
have been awarded) | AO1.1b | A1F
Uses initial conditions to find
arbitrary constants for ‘their’
solution | AO1.1a | M1
Obtains correct value for one of
‘their’ constants (ft only if all M1
marks have been awarded) | AO1.1b | A1F
Obtains correct value for both of
‘their’ constants (ft only if all M1
marks have been awarded) | AO1.1b | A1F
Uses ‘their’ model to describe the
motion of the particle either as a
written description or shown on a
clearly labelled graph. | AO3.4 | A1F
Q | Marking Instructions | AO | Marks | Typical Solution
(b) | Refines their DE model to account
for altered resistive force by
introducing a new coefficient for
dx
dt | AO3.5c | B1 | d2x dx
+α +8x =0
dt2 dt
Critical damping ⇒
λ2 +αλ+8=0 must have equal
roots
α2 =32
α= 4 2
Resistive force should have
magnitude 4 2Mv
Uses or states condition for critical
damping | AO1.2 | B1
Deduces value for coefficient of
dx
dt | AO2.2a | R1
States resistive force | AO3.4 | B1
Total | 12
Q | Marking Instructions | AO | Marks | Typical Solution