OCR MEI Further Mechanics Major 2021 November — Question 12 18 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2021
SessionNovember
Marks18
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Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeString becomes slack timing
DifficultyChallenging +1.2 This is a structured multi-part SHM question with clear guidance at each step. Part (a) uses energy conservation (standard technique), part (b) derives a forced SHM equation from Newton's second law (routine for Further Maths), and part (c) applies a given substitution to solve the differential equation. While it requires multiple techniques and careful algebra, the question provides substantial scaffolding and uses standard Further Mechanics methods without requiring novel insight.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x4.10g Damped oscillations: model and interpret6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle
12 A particle P of mass \(m\) is fixed to one end of a light elastic string of natural length \(l\) and modulus of elasticity 12 mg . The other end of the string is attached to a fixed point O . Particle P is held next to O and then released from rest.
  1. Show that P next comes instantaneously to rest when the length of the string is \(\frac { 3 } { 2 } l\). The string first becomes taut at time \(t = 0\). At time \(t \geqslant 0\), the length of the string is \(l + x\), where \(x\) is the extension in the string.
  2. Show that when the string is taut, \(x\) satisfies the differential equation $$\ddot { \mathrm { x } } + \omega ^ { 2 } \mathrm { x } = \mathrm { g } \text {, where } \omega ^ { 2 } = \frac { 12 \mathrm {~g} } { \mathrm { I } } \text {. }$$
  3. By using the substitution \(x = y + \frac { g } { \omega ^ { 2 } }\), solve the differential equation to show that the time when the string first becomes slack satisfies the equation $$\cos \omega \mathrm { t } - \sqrt { \mathrm { k } } \sin \omega \mathrm { t } = 1$$ where \(k\) is an integer to be determined.
Question 12:
AnswerMarks Guidance
12(a) PE = −mg (l + e) (while P is at rest)
12mge2
EPE =
2l
6mge2
− mg (l + e )= 0
l
6e2 − el − l2 = 0
(3e + l )(2e − l ) = 0
l 1 3
e = ⇒ length of string is l + l = l
AnswerMarks
2 2 2B1
B1
M1*
M1dep*
A1
AnswerMarks
[5]1.1
1.1
3.3
1.1a
AnswerMarks
2.2aWhere e is the extension in the string
Conservation of energy with correct
number of terms
Solving three-term quadratic in e
AnswerMarks
AGTaking the horizontal
through O as the
reference level for
zero GPE
AnswerMarks Guidance
12(b) mg − T = mx
12mgx
mg − = mx
l
12g 12g
x + x = g so x + ω2 x = g where ω2 =
AnswerMarks
l lM1
M1
A1
AnswerMarks
[3]3.3
3.4
AnswerMarks
2.2aN2L vertically with correct number of
terms
Use of Hooke’s law and substitute for T
in N2L
AG
AnswerMarks Guidance
12(c) g
x = y + ⇒ y + ω 2 y = 0
ω 2
y = Acosωt + B sinωt
g
x = Acosωt + B sin ωt +
ω2
g
t = 0, x = 0 ⇒ A = −
ω 2
1
mv 2 = mgl
2
P
v = 2gl
P
2gl
t = 0, x = 2gl ⇒ B =
ω
g 2gl g
x = − cosωt + sin ωt +
ω2 ω ω2
l ( )
1 − cosωt + 24 sin ωt = 0
12
AnswerMarks
cosωt − 24 sinωt =1 so k = 24M1
A1ft
A1
M1
M1*
A1
M1dep*
A1
M1
A1
AnswerMarks
[10]1.1
1.2
1.1
3.4
3.1b
1.1
3.4
1.1
3.1b
AnswerMarks
2.2aUse given substitution to form
differential equation in y
Correctly solves their differential
equation in y
l
oe e.g. x = Acosωt + B sin ωt +
12
Use correct initial conditions in their
expression for x
Use conservation of energy to find speed
v of P at time t = 0
P
Use initial speed in an expression for x
l ( )
oe e.g. x = 1 − cosωt + 24 sin ωt
12
12g
Sets x = 0 and replaces ω2 =
l
AnswerMarks
k need not be stated explicitlyDependent on all
previous M marks
PMT
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Question 12:
12 | (a) | PE = −mg (l + e) (while P is at rest)
12mge2
EPE =
2l
6mge2
− mg (l + e )= 0
l
6e2 − el − l2 = 0
(3e + l )(2e − l ) = 0
l 1 3
e = ⇒ length of string is l + l = l
2 2 2 | B1
B1
M1*
M1dep*
A1
[5] | 1.1
1.1
3.3
1.1a
2.2a | Where e is the extension in the string
Conservation of energy with correct
number of terms
Solving three-term quadratic in e
AG | Taking the horizontal
through O as the
reference level for
zero GPE
12 | (b) | mg − T = mx
12mgx
mg − = mx
l
12g 12g
x + x = g so x + ω2 x = g where ω2 =
l l | M1
M1
A1
[3] | 3.3
3.4
2.2a | N2L vertically with correct number of
terms
Use of Hooke’s law and substitute for T
in N2L
AG
12 | (c) | g
x = y + ⇒ y + ω 2 y = 0
ω 2
y = Acosωt + B sinωt
g
x = Acosωt + B sin ωt +
ω2
g
t = 0, x = 0 ⇒ A = −
ω 2
1
mv 2 = mgl
2
P
v = 2gl
P
2gl
t = 0, x = 2gl ⇒ B =
ω
g 2gl g
x = − cosωt + sin ωt +
ω2 ω ω2
l ( )
1 − cosωt + 24 sin ωt = 0
12
cosωt − 24 sinωt =1 so k = 24 | M1
A1ft
A1
M1
M1*
A1
M1dep*
A1
M1
A1
[10] | 1.1
1.2
1.1
3.4
3.1b
1.1
3.4
1.1
3.1b
2.2a | Use given substitution to form
differential equation in y
Correctly solves their differential
equation in y
l
oe e.g. x = Acosωt + B sin ωt +
12
Use correct initial conditions in their
expression for x
Use conservation of energy to find speed
v of P at time t = 0
P
Use initial speed in an expression for x
l ( )
oe e.g. x = 1 − cosωt + 24 sin ωt
12
12g
Sets x = 0 and replaces ω2 =
l
k need not be stated explicitly | Dependent on all
previous M marks
PMT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored
12 A particle P of mass $m$ is fixed to one end of a light elastic string of natural length $l$ and modulus of elasticity 12 mg . The other end of the string is attached to a fixed point O . Particle P is held next to O and then released from rest.
\begin{enumerate}[label=(\alph*)]
\item Show that P next comes instantaneously to rest when the length of the string is $\frac { 3 } { 2 } l$.

The string first becomes taut at time $t = 0$. At time $t \geqslant 0$, the length of the string is $l + x$, where $x$ is the extension in the string.
\item Show that when the string is taut, $x$ satisfies the differential equation

$$\ddot { \mathrm { x } } + \omega ^ { 2 } \mathrm { x } = \mathrm { g } \text {, where } \omega ^ { 2 } = \frac { 12 \mathrm {~g} } { \mathrm { I } } \text {. }$$
\item By using the substitution $x = y + \frac { g } { \omega ^ { 2 } }$, solve the differential equation to show that the time when the string first becomes slack satisfies the equation

$$\cos \omega \mathrm { t } - \sqrt { \mathrm { k } } \sin \omega \mathrm { t } = 1$$

where $k$ is an integer to be determined.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2021 Q12 [18]}}
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