Question 13:
13 | (a) | T =0.15g
13.5e
=0.15g⇒e=...
0.45
OA=0.45+e=0.5(m) | M1
M1
A1
[3] | 1.1
3.3
1.1 | Resolving vertically for P at A
Apply Hooke’s law and solve for e – where e is the
extension of the string from its natural length to A
AG - if g = 9.8 used then A0
(b) | d2x dx
0.15 =0.15g−T −0.6
dt2 x dt
d2x 13.5 dx
0.15 =0.15g− (0.05+x)−0.6
dt2 0.45 dt
d2x dx
0.15 =−0.6 −30x
dt2 dt
d2x dx
⇒ +4 +200x=0
dt2 dt | M1*
M1dep*
A1
[3] | 2.1
1.1
2.2a | Apply N2L vertically – correct number of terms
Substitute correct expression for the tension when the
extension is 0.05 + x
AG if g = 9.8 used then A0
(c) | 5 d2x dx
x= e −2tsin(14t) and + 4 + 200x= 0
56 dt2 dt
dx 5 5
= e −2tcos(14t)− e −2tsin(14t)
dt 4 28
dx 5 5
t =0, = e −0cos(0)− e −0sin(0)=1.25
dt 4 28
5
t =0,x= e −0sin(0)=0
56
d2x 120
=− e −2tsin(14t)−5e −2tcos(14t)
dt2 7
2
𝑑𝑑 𝑥𝑥 𝑑𝑑𝑥𝑥
2 +4 +200𝑥𝑥
𝑑𝑑𝑡𝑡 𝑑𝑑𝑡𝑡
−2𝑡𝑡 120
= e �− 𝑠𝑠𝑠𝑠𝑠𝑠(14𝑡𝑡)−5𝑐𝑐𝑐𝑐𝑠𝑠(14𝑡𝑡)�
7
−2𝑡𝑡 5 5
+4𝑒𝑒 � 𝑐𝑐𝑐𝑐𝑠𝑠(14𝑡𝑡)− 𝑠𝑠𝑠𝑠𝑠𝑠(14𝑡𝑡)�
4 28
−2𝑡𝑡 5
=e −2tsin(14t)   − 120 − 20+ + 1200000𝑒𝑒  +e �−2 5 t 6co 𝑠𝑠 s 𝑠𝑠𝑠𝑠(1 ( 4 1 t 4)𝑡𝑡() − � 5+5)=0
 7 28 56  | M1*
A1
B1
A1
M1dep*
A1
[6] | 1.1
1.1
3.4
1.1
3.4
2.2a | Differentiating x using the product rule (two terms of
the form uv¢+ vu¢with sin(14t)® ± cos(14t))
dx 5
= e −2t( 14cos(14t)−2sin(14t))
dt 56
dx
Verifying that when t = 0, =1.25and x = 0 – from
dt
a correct first derivative
Correct second derivative (need not be simplified),
for example
d2x = 5 ( −2e −2t )( 14cos(14t)−2sin(14t))
dt2 56
5
+ e −2t(−196sin(14t)−28cos(14t))
56
Substitute x, and its first and second derivatives into
the correct differential equation
AG – sufficient working must be shown
Alternative method (solving the given differential equation) | Solving the differential equation
d2x dx
+ 4 + 200x= 0
dt2 dt
m2+ 4m+ 200= 0Þ m= - 2± 14i | B1 | Correct roots of auxiliary equation – allow - 2+14i
if correct general solution seen
x= e- 2t(Acos(14t)+ Bsin(14t)) | M1* | Correct general solution from their roots of the
auxiliary equation
t= 0,x= 0Þ A= 0 | A1 | Must follow from correct general solution | Must follow from correct general solution
dx
= e- 2t(14Bcos(14t))- 2e- 2t(Bsin(14t))
dt | M1dep* | Differentiating using the product rule (ignore terms in
A)
dx 5
t= 0, = 1.25Þ B=
dt 56 | A1 | Dependent on correctly differentiated terms in B
5
x= e −2tsin(14t)
56 | A1 | Dependent on completely correct working
[6]
(d) | dx 5
= (e- 2t(14cos(14t))- 2e- 2t(sin(14t)))= 0
dt 56
tan(14t)=7
t =0.10206...,0.32646...
when t = 0.32646…, x=−0.046007...
−0.046007... <0.05⇒string does not go slack | M1*
A1
M1dep*
M1
A1
[5] | 3.1b
1.1
3.1a
3.4
2.2b | Setting first derivative (two terms from the product
rule) equal to zero (to find when P is at rest)
Or equivalent 5 e −2t(7cos14t−sin14t)=0
28
Either finding the correct first two times from their
tan(14t)=k for any non-zero k or just considers the
second time (stating 0.326… is sufficient for this
mark)
Finding the displacement of P when P is at rest for the
second time – dependent on both previous M marks
Comparison and conclusion that the string does not
go slack
Differentiating using the product rule (ignore terms in
A)