Question 11:
--- 11(a) ---
11(a) | Forms equilibrium force
equation with three
correct terms.
Condone sign errors | 3.1b | B1 | In equilibrium position
7𝑒𝑒𝐴𝐴 = 9𝑒𝑒𝐵𝐵+0.32𝑔𝑔sin30
,
1.2= 𝑒𝑒𝐴𝐴 +𝑒𝑒𝐵𝐵
𝑒𝑒𝐴𝐴 = 0.775 𝑒𝑒𝐵𝐵 = 0.425
After release
9(𝑒𝑒𝐵𝐵 −𝑥𝑥)+0.32𝑔𝑔sin30−7(𝑒𝑒𝐴𝐴+𝑥𝑥)= 0.32𝑥𝑥̈
9(0.425−𝑥𝑥)+0.32𝑔𝑔sin30−7(0.775+𝑥𝑥)= 0.32𝑥𝑥̈
3.825−9𝑥𝑥+1.6−5.425−7𝑥𝑥 =0.32𝑥𝑥̈
as required −16𝑥𝑥 = 0.32𝑥𝑥̈
𝑥𝑥̈ +50𝑥𝑥 = 0
Forms at least one
correct expression in x
for the tension
ie or
Condone their incorrect
9o(r𝑒𝑒 𝐵𝐵 − 𝑥𝑥) 7(𝑒𝑒𝐴𝐴+𝑥𝑥) | 1.1a | B1F
𝑒𝑒 𝐴𝐴 𝑒𝑒𝐵𝐵
Forms general force
equation with four terms
(with at least two terms
correct).
Condone “a” for
x
Condone sign errors on
the terms
Condone their incorrect
or | 3.1b | M1
𝑒𝑒𝐴𝐴 𝑒𝑒𝐵𝐵
Forms correct force
equation.
Can be in terms of
&
Condone “a” for
x
𝑒𝑒C 𝐴𝐴 ond𝑒𝑒o 𝐵𝐵 ne their incorrect
or | 1.1b | A1F
𝑒𝑒𝐴𝐴 𝑒𝑒𝐵𝐵
Constructs a rigorous
argument to show the
required result | 2.1 | R1
Total | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 11(b)(i) ---
11(b)(i) | Obtains fully correct 2nd
order DE or auxiliary
equation, any correct
form.
Condone −k
Allow instead of 0.32
m | 2.2a | B1 | 9(0.425−x)+0.32gsin30−7(0.775+x)−kx=0.32x
0.32x+kx+16x=0
0.32λ2+kλ+16=0
Critical Damping so:
b2 −4ac=0
k2 −4×0.32×16=0
512
k2 =
25
16 2
k =
5
Sets up b2 −4ac =0
from their 2nd order DE
or Auxiliary Equation | 1.2 | M1
Obtains correct value of
k from correct working | 2.1 | R1
Total | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 11(b)(ii) ---
11(b)(ii) | Obtains correct solution
from their three term
Auxiliary Equation | 3.1a | M1 | 16 2
0.32λ2 + λ+16=0
5
λ=−5 2 twice
x= Ae−5 2 t +Bte−5 2 t
x=0.2,t =0⇒ A=0.2
x =−5 2Ae−5 2 t +Be−5 2 t −5 2Bte−5 2 t
x =0,t =0⇒ B = 2
x=0.2e−5 2 t + 2te−5 2 t
Obtains correct solution
of their three term
differential equation | 1.1b | A1F
Uses x =0.2when
t =0to obtain correct
A | 3.3 | B1
Sets their correct x =0
when t =0 | 3.3 | M1
Obtains their correct
B
16 2
Ft from λ2+ λ+50=0
5 | 1.1b | A1F
Completes a rigorous
argument to obtain the
correct result | 2.1 | R1
Total | 6
Q | Marking instructions | AO | Marks | Typical solution
--- 11(b)(iii) ---
11(b)(iii) | Forms an equation to
find the value of t at the
maximum speed
Must start from a
damped harmonic
model | 3.1a | M1 | x =− 2e−5 2 t + 2e−5 2 t −10te−5 2 t
=−10te−5 2 t
x=−10e−5 2 t +50 2te−5 2 t
At Max
x
x=0=−10e−5 2 t +50 2te−5 2 t
5 2t =1
1 2
t = =
5 2 10
x =− 2e−1 =−0.52026
Max Speed = 0.5 m s-1
Obtains a correct
equation to find the
value of t at the
maximum speed | 1.1b | A1F
Obtains their correct ,
t
any form | 1.1b | A1F
Uses their value of t to
obtain the velocity
Must start from a
damped harmonic
model | 3.4 | M1
Obtains correct max
speed, to at least 1sf or
exact
Condone missing units
(–0.5 m s-1 = A0) | 3.2a | A1
Total | 5
Question total | 19
Q | Marking instructions | AO | Marks | Typical solution