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SHM and Projectile Motion
HSC Revision DaySimple Harmonic Motion and Projectile Motion
Woonona High School
2017
Teacher: Paul HancockSchool: Woonona High SchoolE-mail: [email protected]
Paul Hancock (Woonona High School) Series T1 2017 1 / 24
SHM and Projectile Motion Review
HSC Question Analysis
Let’s have a quick look at the number of marks allocated to the topic for HSCexams in the current format.
Year SHM Marks Projectile Marks Combined marks2012 6 6 122013 2 5 72014 3 10 132015 6 5 112016 5 7 12
The average number of marks per exam is 11.
Paul Hancock (Woonona High School) Series T1 2017 2 / 24
SHM and Projectile Motion Review
HSC Question Analysis
Let’s have a quick look at the number of marks allocated to the topic for HSCexams in the current format.
Year SHM Marks Projectile Marks Combined marks2012 6 6 122013 2 5 72014 3 10 132015 6 5 112016 5 7 12
The average number of marks per exam is 11.
Paul Hancock (Woonona High School) Series T1 2017 2 / 24
SHM and Projectile Motion Review
HSC Question Analysis
Let’s have a quick look at the number of marks allocated to the topic for HSCexams in the current format.
Year SHM Marks Projectile Marks Combined marks2012 6 6 122013 2 5 72014 3 10 132015 6 5 112016 5 7 12
The average number of marks per exam is 11.
Paul Hancock (Woonona High School) Series T1 2017 2 / 24
SHM and Projectile Motion Review
The Syllabus
The Mathematics Syllabus can be found at:
http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/maths23u_syl.pdf
The pages relevant to us today are 76-77. Let’s take a look at some importantpieces of information from the syllabus.
Paul Hancock (Woonona High School) Series T1 2017 3 / 24
SHM and Projectile Motion Review
Simple Harmonic Motion
"Graphs of x , x and x as functions of t should be sketched and therelationships between zero, minimum and maximum values of the threequantities noted. The physical significance of the parameters a, n and αshould be understood, as should the terms amplitude, frequency, period andphase."
You should be very familiar with the equations:
I x = a sin(nt + α) or x = a cos(nt + α)I v = an cos(nt + α) or v = −an sin(nt + α)I v2 = −n2(x2 − a2)I dv
dt= dv
dxdxdt
= v dvdx
I dvdt
= ddx
( 12v
2)
Paul Hancock (Woonona High School) Series T1 2017 4 / 24
SHM and Projectile Motion Review
Simple Harmonic Motion
"Graphs of x , x and x as functions of t should be sketched and therelationships between zero, minimum and maximum values of the threequantities noted. The physical significance of the parameters a, n and αshould be understood, as should the terms amplitude, frequency, period andphase."You should be very familiar with the equations:
I x = a sin(nt + α) or x = a cos(nt + α)I v = an cos(nt + α) or v = −an sin(nt + α)I v2 = −n2(x2 − a2)I dv
dt= dv
dxdxdt
= v dvdx
I dvdt
= ddx
( 12v
2)
Paul Hancock (Woonona High School) Series T1 2017 4 / 24
SHM and Projectile Motion Review
Simple Harmonic Motion
"Graphs of x , x and x as functions of t should be sketched and therelationships between zero, minimum and maximum values of the threequantities noted. The physical significance of the parameters a, n and αshould be understood, as should the terms amplitude, frequency, period andphase."You should be very familiar with the equations:
I x = a sin(nt + α) or x = a cos(nt + α)I v = an cos(nt + α) or v = −an sin(nt + α)I v2 = −n2(x2 − a2)I dv
dt= dv
dxdxdt
= v dvdx
I dvdt
= ddx
( 12v
2)
Paul Hancock (Woonona High School) Series T1 2017 4 / 24
SHM and Projectile Motion Review
Simple Harmonic Motion
"Graphs of x , x and x as functions of t should be sketched and therelationships between zero, minimum and maximum values of the threequantities noted. The physical significance of the parameters a, n and αshould be understood, as should the terms amplitude, frequency, period andphase."You should be very familiar with the equations:
I x = a sin(nt + α) or x = a cos(nt + α)
I v = an cos(nt + α) or v = −an sin(nt + α)I v2 = −n2(x2 − a2)I dv
dt= dv
dxdxdt
= v dvdx
I dvdt
= ddx
( 12v
2)
Paul Hancock (Woonona High School) Series T1 2017 4 / 24
SHM and Projectile Motion Review
Simple Harmonic Motion
"Graphs of x , x and x as functions of t should be sketched and therelationships between zero, minimum and maximum values of the threequantities noted. The physical significance of the parameters a, n and αshould be understood, as should the terms amplitude, frequency, period andphase."You should be very familiar with the equations:
I x = a sin(nt + α) or x = a cos(nt + α)I v = an cos(nt + α) or v = −an sin(nt + α)
I v2 = −n2(x2 − a2)I dv
dt= dv
dxdxdt
= v dvdx
I dvdt
= ddx
( 12v
2)
Paul Hancock (Woonona High School) Series T1 2017 4 / 24
SHM and Projectile Motion Review
Simple Harmonic Motion
"Graphs of x , x and x as functions of t should be sketched and therelationships between zero, minimum and maximum values of the threequantities noted. The physical significance of the parameters a, n and αshould be understood, as should the terms amplitude, frequency, period andphase."You should be very familiar with the equations:
I x = a sin(nt + α) or x = a cos(nt + α)I v = an cos(nt + α) or v = −an sin(nt + α)I v2 = −n2(x2 − a2)
I dvdt
= dvdx
dxdt
= v dvdx
I dvdt
= ddx
( 12v
2)
Paul Hancock (Woonona High School) Series T1 2017 4 / 24
SHM and Projectile Motion Review
Simple Harmonic Motion
"Graphs of x , x and x as functions of t should be sketched and therelationships between zero, minimum and maximum values of the threequantities noted. The physical significance of the parameters a, n and αshould be understood, as should the terms amplitude, frequency, period andphase."You should be very familiar with the equations:
I x = a sin(nt + α) or x = a cos(nt + α)I v = an cos(nt + α) or v = −an sin(nt + α)I v2 = −n2(x2 − a2)I dv
dt= dv
dxdxdt
= v dvdx
I dvdt
= ddx
( 12v
2)
Paul Hancock (Woonona High School) Series T1 2017 4 / 24
SHM and Projectile Motion Review
Simple Harmonic Motion
"Graphs of x , x and x as functions of t should be sketched and therelationships between zero, minimum and maximum values of the threequantities noted. The physical significance of the parameters a, n and αshould be understood, as should the terms amplitude, frequency, period andphase."You should be very familiar with the equations:
I x = a sin(nt + α) or x = a cos(nt + α)I v = an cos(nt + α) or v = −an sin(nt + α)I v2 = −n2(x2 − a2)I dv
dt= dv
dxdxdt
= v dvdx
I dvdt
= ddx
( 12v
2)
Paul Hancock (Woonona High School) Series T1 2017 4 / 24
SHM and Projectile Motion Review
Projectile Motion
"The equations of motion of a particle projected vertically upwards should bederived." (My emphasis).
Most questions will probably ask you to derive the equations of motion alongthe way.
Paul Hancock (Woonona High School) Series T1 2017 5 / 24
SHM and Projectile Motion Review
Projectile Motion
"The equations of motion of a particle projected vertically upwards should bederived." (My emphasis).Most questions will probably ask you to derive the equations of motion alongthe way.
Paul Hancock (Woonona High School) Series T1 2017 5 / 24
SHM and Projectile Motion Multiple Choice
2012 - Question 6Example 1 (Q6)
Well, we have an equation in the form v2 = −n2(x2 − a2), all that is required is alittle rearrangement of the terms.v2 = −42(x2 − 32), so n = 4 and A = 3.Now, T is the period and is given by T = 2π
n = 2π4 = π
2
Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 6 / 24
SHM and Projectile Motion Multiple Choice
2012 - Question 6Example 1 (Q6)
Well, we have an equation in the form v2 = −n2(x2 − a2), all that is required is alittle rearrangement of the terms.
v2 = −42(x2 − 32), so n = 4 and A = 3.Now, T is the period and is given by T = 2π
n = 2π4 = π
2
Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 6 / 24
SHM and Projectile Motion Multiple Choice
2012 - Question 6Example 1 (Q6)
Well, we have an equation in the form v2 = −n2(x2 − a2), all that is required is alittle rearrangement of the terms.v2 = −42(x2 − 32), so n = 4 and A = 3.
Now, T is the period and is given by T = 2πn = 2π
4 = π2
Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 6 / 24
SHM and Projectile Motion Multiple Choice
2012 - Question 6Example 1 (Q6)
Well, we have an equation in the form v2 = −n2(x2 − a2), all that is required is alittle rearrangement of the terms.v2 = −42(x2 − 32), so n = 4 and A = 3.Now, T is the period and is given by T = 2π
n = 2π4 = π
2 Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 6 / 24
SHM and Projectile Motion Multiple Choice
2012 - Question 6Example 1 (Q6)
Well, we have an equation in the form v2 = −n2(x2 − a2), all that is required is alittle rearrangement of the terms.v2 = −42(x2 − 32), so n = 4 and A = 3.Now, T is the period and is given by T = 2π
n = 2π4 = π
2 Answer: (A)
Paul Hancock (Woonona High School) Series T1 2017 6 / 24
SHM and Projectile Motion Multiple Choice
2014 - Question 7Example 2 (Q7)
Well, we have a = 5 and T = 6, so from our earlier formula: n = 2π6 = π
3 .We know v must be of the form v = an cos(nt). So, v = 5π
3 cos(π3 t)
Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 7 / 24
SHM and Projectile Motion Multiple Choice
2014 - Question 7Example 2 (Q7)
Well, we have a = 5 and T = 6, so from our earlier formula: n = 2π6 = π
3 .
We know v must be of the form v = an cos(nt). So, v = 5π3 cos(π3 t)
Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 7 / 24
SHM and Projectile Motion Multiple Choice
2014 - Question 7Example 2 (Q7)
Well, we have a = 5 and T = 6, so from our earlier formula: n = 2π6 = π
3 .We know v must be of the form v = an cos(nt). So, v = 5π
3 cos(π3 t)
Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 7 / 24
SHM and Projectile Motion Multiple Choice
2014 - Question 7Example 2 (Q7)
Well, we have a = 5 and T = 6, so from our earlier formula: n = 2π6 = π
3 .We know v must be of the form v = an cos(nt). So, v = 5π
3 cos(π3 t)
Answer: (A)
Paul Hancock (Woonona High School) Series T1 2017 7 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 12a
Paul Hancock (Woonona High School) Series T1 2017 8 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 12a
(i) What is the total distance travelled by the particle when it first returns to theorigin?
Well at time t = 0 the particle is at the origin. So it first returns to the originafter travelling outwards a distance equal to the amplitude and then returninginwards the same distance.So we simply need to double the amplitude.From the question the amplitude is 2 m, so our final solution is 4 m.
Paul Hancock (Woonona High School) Series T1 2017 9 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 12a
(i) What is the total distance travelled by the particle when it first returns to theorigin?
Well at time t = 0 the particle is at the origin. So it first returns to the originafter travelling outwards a distance equal to the amplitude and then returninginwards the same distance.
So we simply need to double the amplitude.From the question the amplitude is 2 m, so our final solution is 4 m.
Paul Hancock (Woonona High School) Series T1 2017 9 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 12a
(i) What is the total distance travelled by the particle when it first returns to theorigin?
Well at time t = 0 the particle is at the origin. So it first returns to the originafter travelling outwards a distance equal to the amplitude and then returninginwards the same distance.So we simply need to double the amplitude.
From the question the amplitude is 2 m, so our final solution is 4 m.
Paul Hancock (Woonona High School) Series T1 2017 9 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 12a
(i) What is the total distance travelled by the particle when it first returns to theorigin?
Well at time t = 0 the particle is at the origin. So it first returns to the originafter travelling outwards a distance equal to the amplitude and then returninginwards the same distance.So we simply need to double the amplitude.From the question the amplitude is 2 m, so our final solution is 4 m.
Paul Hancock (Woonona High School) Series T1 2017 9 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 12a
(ii) What is the acceleration of the particle when it is first at rest?
Well, we need to know when the particle is at rest...This occurs when the particle has travelled to maximum displacement, ie.when it is a distance from the origin equal to its displacement.
2 sin 3t = 2sin 3t = 1
3t =π
2t =
π
6
Paul Hancock (Woonona High School) Series T1 2017 10 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 12a
(ii) What is the acceleration of the particle when it is first at rest?
Well, we need to know when the particle is at rest...
This occurs when the particle has travelled to maximum displacement, ie.when it is a distance from the origin equal to its displacement.
2 sin 3t = 2sin 3t = 1
3t =π
2t =
π
6
Paul Hancock (Woonona High School) Series T1 2017 10 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 12a
(ii) What is the acceleration of the particle when it is first at rest?
Well, we need to know when the particle is at rest...This occurs when the particle has travelled to maximum displacement, ie.when it is a distance from the origin equal to its displacement.
2 sin 3t = 2sin 3t = 1
3t =π
2t =
π
6
Paul Hancock (Woonona High School) Series T1 2017 10 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 12a
(ii) What is the acceleration of the particle when it is first at rest?
Well, we need to know when the particle is at rest...This occurs when the particle has travelled to maximum displacement, ie.when it is a distance from the origin equal to its displacement.
2 sin 3t = 2
sin 3t = 1
3t =π
2t =
π
6
Paul Hancock (Woonona High School) Series T1 2017 10 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 12a
(ii) What is the acceleration of the particle when it is first at rest?
Well, we need to know when the particle is at rest...This occurs when the particle has travelled to maximum displacement, ie.when it is a distance from the origin equal to its displacement.
2 sin 3t = 2sin 3t = 1
3t =π
2t =
π
6
Paul Hancock (Woonona High School) Series T1 2017 10 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 12a
(ii) What is the acceleration of the particle when it is first at rest?
Well, we need to know when the particle is at rest...This occurs when the particle has travelled to maximum displacement, ie.when it is a distance from the origin equal to its displacement.
2 sin 3t = 2sin 3t = 1
3t =π
2
t =π
6
Paul Hancock (Woonona High School) Series T1 2017 10 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 12a
(ii) What is the acceleration of the particle when it is first at rest?
Well, we need to know when the particle is at rest...This occurs when the particle has travelled to maximum displacement, ie.when it is a distance from the origin equal to its displacement.
2 sin 3t = 2sin 3t = 1
3t =π
2t =
π
6
Paul Hancock (Woonona High School) Series T1 2017 10 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 12e
We can solve this problem in 2 ways.
v2 = −9(x2 − k
9
)
= −9
x2 −
(√k
3
)2
hence, n = 3 and so T = 2π3
.
x =d
dx
(12v2)
=d
dx
(k
2− 9
2x2)
= −9x
hence, n = 3 and so T = 2π3
.
Paul Hancock (Woonona High School) Series T1 2017 11 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 12e
We can solve this problem in 2 ways.
v2 = −9(x2 − k
9
)
= −9
x2 −
(√k
3
)2
hence, n = 3 and so T = 2π3
.
x =d
dx
(12v2)
=d
dx
(k
2− 9
2x2)
= −9x
hence, n = 3 and so T = 2π3
.
Paul Hancock (Woonona High School) Series T1 2017 11 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 12e
We can solve this problem in 2 ways.
v2 = −9(x2 − k
9
)
= −9
x2 −
(√k
3
)2
hence, n = 3 and so T = 2π3
.
x =d
dx
(12v2)
=d
dx
(k
2− 9
2x2)
= −9x
hence, n = 3 and so T = 2π3
.
Paul Hancock (Woonona High School) Series T1 2017 11 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 12e
We can solve this problem in 2 ways.
v2 = −9(x2 − k
9
)
= −9
x2 −
(√k
3
)2
hence, n = 3 and so T = 2π3
.
x =d
dx
(12v2)
=d
dx
(k
2− 9
2x2)
= −9x
hence, n = 3 and so T = 2π3
.
Paul Hancock (Woonona High School) Series T1 2017 11 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 12e
We can solve this problem in 2 ways.
v2 = −9(x2 − k
9
)
= −9
x2 −
(√k
3
)2
hence, n = 3 and so T = 2π3 .
x =d
dx
(12v2)
=d
dx
(k
2− 9
2x2)
= −9x
hence, n = 3 and so T = 2π3
.
Paul Hancock (Woonona High School) Series T1 2017 11 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 12e
We can solve this problem in 2 ways.
v2 = −9(x2 − k
9
)
= −9
x2 −
(√k
3
)2
hence, n = 3 and so T = 2π3 .
x =d
dx
(12v2)
=d
dx
(k
2− 9
2x2)
= −9x
hence, n = 3 and so T = 2π3
.
Paul Hancock (Woonona High School) Series T1 2017 11 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 12e
We can solve this problem in 2 ways.
v2 = −9(x2 − k
9
)
= −9
x2 −
(√k
3
)2
hence, n = 3 and so T = 2π3 .
x =d
dx
(12v2)
=d
dx
(k
2− 9
2x2)
= −9x
hence, n = 3 and so T = 2π3
.
Paul Hancock (Woonona High School) Series T1 2017 11 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 12e
We can solve this problem in 2 ways.
v2 = −9(x2 − k
9
)
= −9
x2 −
(√k
3
)2
hence, n = 3 and so T = 2π3 .
x =d
dx
(12v2)
=d
dx
(k
2− 9
2x2)
= −9x
hence, n = 3 and so T = 2π3
.
Paul Hancock (Woonona High School) Series T1 2017 11 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 12e
We can solve this problem in 2 ways.
v2 = −9(x2 − k
9
)
= −9
x2 −
(√k
3
)2
hence, n = 3 and so T = 2π3 .
x =d
dx
(12v2)
=d
dx
(k
2− 9
2x2)
= −9x
hence, n = 3 and so T = 2π3 .
Paul Hancock (Woonona High School) Series T1 2017 11 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
Paul Hancock (Woonona High School) Series T1 2017 12 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(i) Prove that the particle is moving in simple harmonic motion by showing that xsatisfies an equation of the form x = −n2(x − c).
Here we are required to use auxillary angles to simplfy the problem.
6 sin 2t + 8 cos 2t = 10 sin(2t + tan−1(3/4))
x = 5+ 10 sin(2t + tan−1(3/4))
x = 20 cos(2t + tan−1(3/4))
x = −40 sin(2t + tan−1(3/4))= −4(x − 5)
= −22(x − 5)
Paul Hancock (Woonona High School) Series T1 2017 13 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(i) Prove that the particle is moving in simple harmonic motion by showing that xsatisfies an equation of the form x = −n2(x − c).
Here we are required to use auxillary angles to simplfy the problem.
6 sin 2t + 8 cos 2t = 10 sin(2t + tan−1(3/4))
x = 5+ 10 sin(2t + tan−1(3/4))
x = 20 cos(2t + tan−1(3/4))
x = −40 sin(2t + tan−1(3/4))= −4(x − 5)
= −22(x − 5)
Paul Hancock (Woonona High School) Series T1 2017 13 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(i) Prove that the particle is moving in simple harmonic motion by showing that xsatisfies an equation of the form x = −n2(x − c).
Here we are required to use auxillary angles to simplfy the problem.
6 sin 2t + 8 cos 2t = 10 sin(2t + tan−1(3/4))
x = 5+ 10 sin(2t + tan−1(3/4))
x = 20 cos(2t + tan−1(3/4))
x = −40 sin(2t + tan−1(3/4))= −4(x − 5)
= −22(x − 5)
Paul Hancock (Woonona High School) Series T1 2017 13 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(i) Prove that the particle is moving in simple harmonic motion by showing that xsatisfies an equation of the form x = −n2(x − c).
Here we are required to use auxillary angles to simplfy the problem.
6 sin 2t + 8 cos 2t = 10 sin(2t + tan−1(3/4))
x = 5+ 10 sin(2t + tan−1(3/4))
x = 20 cos(2t + tan−1(3/4))
x = −40 sin(2t + tan−1(3/4))= −4(x − 5)
= −22(x − 5)
Paul Hancock (Woonona High School) Series T1 2017 13 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(i) Prove that the particle is moving in simple harmonic motion by showing that xsatisfies an equation of the form x = −n2(x − c).
Here we are required to use auxillary angles to simplfy the problem.
6 sin 2t + 8 cos 2t = 10 sin(2t + tan−1(3/4))
x = 5+ 10 sin(2t + tan−1(3/4))
x = 20 cos(2t + tan−1(3/4))
x = −40 sin(2t + tan−1(3/4))= −4(x − 5)
= −22(x − 5)
Paul Hancock (Woonona High School) Series T1 2017 13 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(i) Prove that the particle is moving in simple harmonic motion by showing that xsatisfies an equation of the form x = −n2(x − c).
Here we are required to use auxillary angles to simplfy the problem.
6 sin 2t + 8 cos 2t = 10 sin(2t + tan−1(3/4))
x = 5+ 10 sin(2t + tan−1(3/4))
x = 20 cos(2t + tan−1(3/4))
x = −40 sin(2t + tan−1(3/4))
= −4(x − 5)
= −22(x − 5)
Paul Hancock (Woonona High School) Series T1 2017 13 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(i) Prove that the particle is moving in simple harmonic motion by showing that xsatisfies an equation of the form x = −n2(x − c).
Here we are required to use auxillary angles to simplfy the problem.
6 sin 2t + 8 cos 2t = 10 sin(2t + tan−1(3/4))
x = 5+ 10 sin(2t + tan−1(3/4))
x = 20 cos(2t + tan−1(3/4))
x = −40 sin(2t + tan−1(3/4))= −4(x − 5)
= −22(x − 5)
Paul Hancock (Woonona High School) Series T1 2017 13 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(i) Prove that the particle is moving in simple harmonic motion by showing that xsatisfies an equation of the form x = −n2(x − c).
Here we are required to use auxillary angles to simplfy the problem.
6 sin 2t + 8 cos 2t = 10 sin(2t + tan−1(3/4))
x = 5+ 10 sin(2t + tan−1(3/4))
x = 20 cos(2t + tan−1(3/4))
x = −40 sin(2t + tan−1(3/4))= −4(x − 5)
= −22(x − 5)
Paul Hancock (Woonona High School) Series T1 2017 13 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(ii) When is the displacement of the particle zero for the first time?
5+ 10 sin(2t + tan−1(3/4)) = 0
sin(2t + tan−1(3/4)) = −12
2t + tan−1(3/4) =7π6
t =7π12− 1
2tan−1(3/4)
Paul Hancock (Woonona High School) Series T1 2017 14 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(ii) When is the displacement of the particle zero for the first time?
5+ 10 sin(2t + tan−1(3/4)) = 0
sin(2t + tan−1(3/4)) = −12
2t + tan−1(3/4) =7π6
t =7π12− 1
2tan−1(3/4)
Paul Hancock (Woonona High School) Series T1 2017 14 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(ii) When is the displacement of the particle zero for the first time?
5+ 10 sin(2t + tan−1(3/4)) = 0
sin(2t + tan−1(3/4)) = −12
2t + tan−1(3/4) =7π6
t =7π12− 1
2tan−1(3/4)
Paul Hancock (Woonona High School) Series T1 2017 14 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(ii) When is the displacement of the particle zero for the first time?
5+ 10 sin(2t + tan−1(3/4)) = 0
sin(2t + tan−1(3/4)) = −12
2t + tan−1(3/4) =7π6
t =7π12− 1
2tan−1(3/4)
Paul Hancock (Woonona High School) Series T1 2017 14 / 24
SHM and Projectile Motion Standard Questions
2012 - Question 13c
(ii) When is the displacement of the particle zero for the first time?
5+ 10 sin(2t + tan−1(3/4)) = 0
sin(2t + tan−1(3/4)) = −12
2t + tan−1(3/4) =7π6
t =7π12− 1
2tan−1(3/4)
Paul Hancock (Woonona High School) Series T1 2017 14 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c
Paul Hancock (Woonona High School) Series T1 2017 15 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c
(i) How long does the projectile fired from A take to reach its maximum height?
Well, firstly we need to find dydt and then set this equal to zero, as the particle
comes to rest in the vertical direction at maximum height.
y = ut sinα− g
2t
dy
dt= u sinα− gt
u sinα− gt = 0
t =u sinα
g
Paul Hancock (Woonona High School) Series T1 2017 16 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c
(i) How long does the projectile fired from A take to reach its maximum height?
Well, firstly we need to find dydt and then set this equal to zero, as the particle
comes to rest in the vertical direction at maximum height.
y = ut sinα− g
2t
dy
dt= u sinα− gt
u sinα− gt = 0
t =u sinα
g
Paul Hancock (Woonona High School) Series T1 2017 16 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c
(i) How long does the projectile fired from A take to reach its maximum height?
Well, firstly we need to find dydt and then set this equal to zero, as the particle
comes to rest in the vertical direction at maximum height.
y = ut sinα− g
2t
dy
dt= u sinα− gt
u sinα− gt = 0
t =u sinα
g
Paul Hancock (Woonona High School) Series T1 2017 16 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c
(i) How long does the projectile fired from A take to reach its maximum height?
Well, firstly we need to find dydt and then set this equal to zero, as the particle
comes to rest in the vertical direction at maximum height.
y = ut sinα− g
2t
dy
dt= u sinα− gt
u sinα− gt = 0
t =u sinα
g
Paul Hancock (Woonona High School) Series T1 2017 16 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c
(i) How long does the projectile fired from A take to reach its maximum height?
Well, firstly we need to find dydt and then set this equal to zero, as the particle
comes to rest in the vertical direction at maximum height.
y = ut sinα− g
2t
dy
dt= u sinα− gt
u sinα− gt = 0
t =u sinα
g
Paul Hancock (Woonona High School) Series T1 2017 16 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c
(ii) Show that u sinα = w sinβ.
Using (i) it should be clear that for particle B, the time to maximum height issimilarly t = w sin β
g .
Since the particles collide at maximum height, then those two times should be thesame.
u sinαg
=w sinβ
g
u sinα = w sinβ
Paul Hancock (Woonona High School) Series T1 2017 17 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c
(ii) Show that u sinα = w sinβ.
Using (i) it should be clear that for particle B, the time to maximum height issimilarly t = w sin β
g .
Since the particles collide at maximum height, then those two times should be thesame.
u sinαg
=w sinβ
g
u sinα = w sinβ
Paul Hancock (Woonona High School) Series T1 2017 17 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c
(ii) Show that u sinα = w sinβ.
Using (i) it should be clear that for particle B, the time to maximum height issimilarly t = w sin β
g .
Since the particles collide at maximum height, then those two times should be thesame.
u sinαg
=w sinβ
g
u sinα = w sinβ
Paul Hancock (Woonona High School) Series T1 2017 17 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c
(ii) Show that u sinα = w sinβ.
Using (i) it should be clear that for particle B, the time to maximum height issimilarly t = w sin β
g .
Since the particles collide at maximum height, then those two times should be thesame.
u sinαg
=w sinβ
g
u sinα = w sinβ
Paul Hancock (Woonona High School) Series T1 2017 17 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c
(ii) Show that u sinα = w sinβ.
Using (i) it should be clear that for particle B, the time to maximum height issimilarly t = w sin β
g .
Since the particles collide at maximum height, then those two times should be thesame.
u sinαg
=w sinβ
g
u sinα = w sinβ
Paul Hancock (Woonona High School) Series T1 2017 17 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c(iii) Show that d = uw
g sin(α+ β).
d is the horizontal distance travelled by A and B from their initial locations to thepoint of collision.
d = u
(u sinα
g
)cosα+ w
(w sinβ
g
)cosβ
=u sinα× u cosα
g+
w sinβ × w cosβg
=w sinβ × u cosα
g+
u sinα× w cosβg
=uw
g(sinβ × cosα+ sinα× cosβ)
=uw
gsin(α+ β)
Paul Hancock (Woonona High School) Series T1 2017 18 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c(iii) Show that d = uw
g sin(α+ β).
d is the horizontal distance travelled by A and B from their initial locations to thepoint of collision.
d = u
(u sinα
g
)cosα+ w
(w sinβ
g
)cosβ
=u sinα× u cosα
g+
w sinβ × w cosβg
=w sinβ × u cosα
g+
u sinα× w cosβg
=uw
g(sinβ × cosα+ sinα× cosβ)
=uw
gsin(α+ β)
Paul Hancock (Woonona High School) Series T1 2017 18 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c(iii) Show that d = uw
g sin(α+ β).
d is the horizontal distance travelled by A and B from their initial locations to thepoint of collision.
d = u
(u sinα
g
)cosα+ w
(w sinβ
g
)cosβ
=u sinα× u cosα
g+
w sinβ × w cosβg
=w sinβ × u cosα
g+
u sinα× w cosβg
=uw
g(sinβ × cosα+ sinα× cosβ)
=uw
gsin(α+ β)
Paul Hancock (Woonona High School) Series T1 2017 18 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c(iii) Show that d = uw
g sin(α+ β).
d is the horizontal distance travelled by A and B from their initial locations to thepoint of collision.
d = u
(u sinα
g
)cosα+ w
(w sinβ
g
)cosβ
=u sinα× u cosα
g+
w sinβ × w cosβg
=w sinβ × u cosα
g+
u sinα× w cosβg
=uw
g(sinβ × cosα+ sinα× cosβ)
=uw
gsin(α+ β)
Paul Hancock (Woonona High School) Series T1 2017 18 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c(iii) Show that d = uw
g sin(α+ β).
d is the horizontal distance travelled by A and B from their initial locations to thepoint of collision.
d = u
(u sinα
g
)cosα+ w
(w sinβ
g
)cosβ
=u sinα× u cosα
g+
w sinβ × w cosβg
=w sinβ × u cosα
g+
u sinα× w cosβg
=uw
g(sinβ × cosα+ sinα× cosβ)
=uw
gsin(α+ β)
Paul Hancock (Woonona High School) Series T1 2017 18 / 24
SHM and Projectile Motion Standard Questions
2013 - Question 13c(iii) Show that d = uw
g sin(α+ β).
d is the horizontal distance travelled by A and B from their initial locations to thepoint of collision.
d = u
(u sinα
g
)cosα+ w
(w sinβ
g
)cosβ
=u sinα× u cosα
g+
w sinβ × w cosβg
=w sinβ × u cosα
g+
u sinα× w cosβg
=uw
g(sinβ × cosα+ sinα× cosβ)
=uw
gsin(α+ β)
Paul Hancock (Woonona High School) Series T1 2017 18 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
Paul Hancock (Woonona High School) Series T1 2017 19 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(i) Show that the Cartesian equation of the flight path of the skier is given byy = x tan θ − gx2
2V 2 sec2 θ.
We need to transform the parametric equations into Cartesian form. We do thisby rearranging the displacement in x in terms of t and substituting into thedisplacement in y .
t =x
V cos θ
y = −12g( x
V cos θ
)2+ V
( x
V cos θ
)sin θ
y = − gx2
2V 2 cos2 θ+ x
sin θcos θ
y = − gx2
2V 2 sec2 θ + x tan θ
Paul Hancock (Woonona High School) Series T1 2017 20 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(i) Show that the Cartesian equation of the flight path of the skier is given byy = x tan θ − gx2
2V 2 sec2 θ.
We need to transform the parametric equations into Cartesian form. We do thisby rearranging the displacement in x in terms of t and substituting into thedisplacement in y .
t =x
V cos θ
y = −12g( x
V cos θ
)2+ V
( x
V cos θ
)sin θ
y = − gx2
2V 2 cos2 θ+ x
sin θcos θ
y = − gx2
2V 2 sec2 θ + x tan θ
Paul Hancock (Woonona High School) Series T1 2017 20 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(i) Show that the Cartesian equation of the flight path of the skier is given byy = x tan θ − gx2
2V 2 sec2 θ.
We need to transform the parametric equations into Cartesian form. We do thisby rearranging the displacement in x in terms of t and substituting into thedisplacement in y .
t =x
V cos θ
y = −12g( x
V cos θ
)2+ V
( x
V cos θ
)sin θ
y = − gx2
2V 2 cos2 θ+ x
sin θcos θ
y = − gx2
2V 2 sec2 θ + x tan θ
Paul Hancock (Woonona High School) Series T1 2017 20 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(i) Show that the Cartesian equation of the flight path of the skier is given byy = x tan θ − gx2
2V 2 sec2 θ.
We need to transform the parametric equations into Cartesian form. We do thisby rearranging the displacement in x in terms of t and substituting into thedisplacement in y .
t =x
V cos θ
y = −12g( x
V cos θ
)2+ V
( x
V cos θ
)sin θ
y = − gx2
2V 2 cos2 θ+ x
sin θcos θ
y = − gx2
2V 2 sec2 θ + x tan θ
Paul Hancock (Woonona High School) Series T1 2017 20 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(i) Show that the Cartesian equation of the flight path of the skier is given byy = x tan θ − gx2
2V 2 sec2 θ.
We need to transform the parametric equations into Cartesian form. We do thisby rearranging the displacement in x in terms of t and substituting into thedisplacement in y .
t =x
V cos θ
y = −12g( x
V cos θ
)2+ V
( x
V cos θ
)sin θ
y = − gx2
2V 2 cos2 θ+ x
sin θcos θ
y = − gx2
2V 2 sec2 θ + x tan θ
Paul Hancock (Woonona High School) Series T1 2017 20 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(i) Show that the Cartesian equation of the flight path of the skier is given byy = x tan θ − gx2
2V 2 sec2 θ.
We need to transform the parametric equations into Cartesian form. We do thisby rearranging the displacement in x in terms of t and substituting into thedisplacement in y .
t =x
V cos θ
y = −12g( x
V cos θ
)2+ V
( x
V cos θ
)sin θ
y = − gx2
2V 2 cos2 θ+ x
sin θcos θ
y = − gx2
2V 2 sec2 θ + x tan θ
Paul Hancock (Woonona High School) Series T1 2017 20 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(ii) Show that D = 2
√2V 2
g cos θ(cos θ + sin θ).
P occurs at the intersection of the skiers trajectory and the line OP. Now, OPhas the equation y = −x , so we must solve simultaneous equations.
x tan θ − gx2
2V 2 sec2 θ = −xg
2V 2 sec2 θx2 − (1+ tan θ)x = 0
This is just a rather scary looking quadratic equation.
x =2V 2(1+ tan θ)
g sec2 θ=
2V 2
g
(1
sec2θ+
tan θsec2 θ
)=
2V 2
g
(cos2 θ + cos θ sin θ
)x =
2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 21 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(ii) Show that D = 2
√2V 2
g cos θ(cos θ + sin θ).
P occurs at the intersection of the skiers trajectory and the line OP. Now, OPhas the equation y = −x , so we must solve simultaneous equations.
x tan θ − gx2
2V 2 sec2 θ = −xg
2V 2 sec2 θx2 − (1+ tan θ)x = 0
This is just a rather scary looking quadratic equation.
x =2V 2(1+ tan θ)
g sec2 θ=
2V 2
g
(1
sec2θ+
tan θsec2 θ
)=
2V 2
g
(cos2 θ + cos θ sin θ
)x =
2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 21 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(ii) Show that D = 2
√2V 2
g cos θ(cos θ + sin θ).
P occurs at the intersection of the skiers trajectory and the line OP. Now, OPhas the equation y = −x , so we must solve simultaneous equations.
x tan θ − gx2
2V 2 sec2 θ = −x
g
2V 2 sec2 θx2 − (1+ tan θ)x = 0
This is just a rather scary looking quadratic equation.
x =2V 2(1+ tan θ)
g sec2 θ=
2V 2
g
(1
sec2θ+
tan θsec2 θ
)=
2V 2
g
(cos2 θ + cos θ sin θ
)x =
2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 21 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(ii) Show that D = 2
√2V 2
g cos θ(cos θ + sin θ).
P occurs at the intersection of the skiers trajectory and the line OP. Now, OPhas the equation y = −x , so we must solve simultaneous equations.
x tan θ − gx2
2V 2 sec2 θ = −xg
2V 2 sec2 θx2 − (1+ tan θ)x = 0
This is just a rather scary looking quadratic equation.
x =2V 2(1+ tan θ)
g sec2 θ=
2V 2
g
(1
sec2θ+
tan θsec2 θ
)=
2V 2
g
(cos2 θ + cos θ sin θ
)x =
2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 21 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(ii) Show that D = 2
√2V 2
g cos θ(cos θ + sin θ).
P occurs at the intersection of the skiers trajectory and the line OP. Now, OPhas the equation y = −x , so we must solve simultaneous equations.
x tan θ − gx2
2V 2 sec2 θ = −xg
2V 2 sec2 θx2 − (1+ tan θ)x = 0
This is just a rather scary looking quadratic equation.
x =2V 2(1+ tan θ)
g sec2 θ=
2V 2
g
(1
sec2θ+
tan θsec2 θ
)=
2V 2
g
(cos2 θ + cos θ sin θ
)x =
2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 21 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(ii) Show that D = 2
√2V 2
g cos θ(cos θ + sin θ).
P occurs at the intersection of the skiers trajectory and the line OP. Now, OPhas the equation y = −x , so we must solve simultaneous equations.
x tan θ − gx2
2V 2 sec2 θ = −xg
2V 2 sec2 θx2 − (1+ tan θ)x = 0
This is just a rather scary looking quadratic equation.
x =2V 2(1+ tan θ)
g sec2 θ
=2V 2
g
(1
sec2θ+
tan θsec2 θ
)=
2V 2
g
(cos2 θ + cos θ sin θ
)x =
2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 21 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(ii) Show that D = 2
√2V 2
g cos θ(cos θ + sin θ).
P occurs at the intersection of the skiers trajectory and the line OP. Now, OPhas the equation y = −x , so we must solve simultaneous equations.
x tan θ − gx2
2V 2 sec2 θ = −xg
2V 2 sec2 θx2 − (1+ tan θ)x = 0
This is just a rather scary looking quadratic equation.
x =2V 2(1+ tan θ)
g sec2 θ=
2V 2
g
(1
sec2θ+
tan θsec2 θ
)
=2V 2
g
(cos2 θ + cos θ sin θ
)x =
2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 21 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(ii) Show that D = 2
√2V 2
g cos θ(cos θ + sin θ).
P occurs at the intersection of the skiers trajectory and the line OP. Now, OPhas the equation y = −x , so we must solve simultaneous equations.
x tan θ − gx2
2V 2 sec2 θ = −xg
2V 2 sec2 θx2 − (1+ tan θ)x = 0
This is just a rather scary looking quadratic equation.
x =2V 2(1+ tan θ)
g sec2 θ=
2V 2
g
(1
sec2θ+
tan θsec2 θ
)=
2V 2
g
(cos2 θ + cos θ sin θ
)
x =2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 21 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a(ii) Show that D = 2
√2V 2
g cos θ(cos θ + sin θ).
P occurs at the intersection of the skiers trajectory and the line OP. Now, OPhas the equation y = −x , so we must solve simultaneous equations.
x tan θ − gx2
2V 2 sec2 θ = −xg
2V 2 sec2 θx2 − (1+ tan θ)x = 0
This is just a rather scary looking quadratic equation.
x =2V 2(1+ tan θ)
g sec2 θ=
2V 2
g
(1
sec2θ+
tan θsec2 θ
)=
2V 2
g
(cos2 θ + cos θ sin θ
)x =
2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 21 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(ii) Show that D = 2√2V 2
g cos θ(cos θ + sin θ).
D =√x2 + y2
=√
x2 + (−x)2
=√2x2
=√2x
= 2√2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 22 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(ii) Show that D = 2√2V 2
g cos θ(cos θ + sin θ).
D =√x2 + y2
=√x2 + (−x)2
=√2x2
=√2x
= 2√2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 22 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(ii) Show that D = 2√2V 2
g cos θ(cos θ + sin θ).
D =√x2 + y2
=√x2 + (−x)2
=√2x2
=√2x
= 2√2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 22 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(ii) Show that D = 2√2V 2
g cos θ(cos θ + sin θ).
D =√x2 + y2
=√x2 + (−x)2
=√2x2
=√2x
= 2√2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 22 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(ii) Show that D = 2√2V 2
g cos θ(cos θ + sin θ).
D =√x2 + y2
=√x2 + (−x)2
=√2x2
=√2x
= 2√2V 2
gcos θ(cos θ + sin θ)
Paul Hancock (Woonona High School) Series T1 2017 22 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(iii) Show that dDdθ = 2
√2V 2
g (cos 2θ − sin 2θ).
This simply requires us to use the product rule on our answer to (ii).
dD
dθ= 2√2V 2
g[(cos θ + sin θ)(−sinθ) + (cos θ)(− sin θ + cos θ)]
= 2√2V 2
g
[− sin θ cos θ − sin2 θ − sin θ cos θ + cos2 θ
]= 2√2V 2
g
[cos2 θ − sin2 θ − 2 sin θ cos θ
]= 2√2V 2
g[cos 2θ − sin 2θ]
Paul Hancock (Woonona High School) Series T1 2017 23 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(iii) Show that dDdθ = 2
√2V 2
g (cos 2θ − sin 2θ).
This simply requires us to use the product rule on our answer to (ii).
dD
dθ= 2√2V 2
g[(cos θ + sin θ)(−sinθ) + (cos θ)(− sin θ + cos θ)]
= 2√2V 2
g
[− sin θ cos θ − sin2 θ − sin θ cos θ + cos2 θ
]= 2√2V 2
g
[cos2 θ − sin2 θ − 2 sin θ cos θ
]= 2√2V 2
g[cos 2θ − sin 2θ]
Paul Hancock (Woonona High School) Series T1 2017 23 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(iii) Show that dDdθ = 2
√2V 2
g (cos 2θ − sin 2θ).
This simply requires us to use the product rule on our answer to (ii).
dD
dθ= 2√2V 2
g[(cos θ + sin θ)(−sinθ) + (cos θ)(− sin θ + cos θ)]
= 2√2V 2
g
[− sin θ cos θ − sin2 θ − sin θ cos θ + cos2 θ
]= 2√2V 2
g
[cos2 θ − sin2 θ − 2 sin θ cos θ
]= 2√2V 2
g[cos 2θ − sin 2θ]
Paul Hancock (Woonona High School) Series T1 2017 23 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(iii) Show that dDdθ = 2
√2V 2
g (cos 2θ − sin 2θ).
This simply requires us to use the product rule on our answer to (ii).
dD
dθ= 2√2V 2
g[(cos θ + sin θ)(−sinθ) + (cos θ)(− sin θ + cos θ)]
= 2√2V 2
g
[− sin θ cos θ − sin2 θ − sin θ cos θ + cos2 θ
]
= 2√2V 2
g
[cos2 θ − sin2 θ − 2 sin θ cos θ
]= 2√2V 2
g[cos 2θ − sin 2θ]
Paul Hancock (Woonona High School) Series T1 2017 23 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(iii) Show that dDdθ = 2
√2V 2
g (cos 2θ − sin 2θ).
This simply requires us to use the product rule on our answer to (ii).
dD
dθ= 2√2V 2
g[(cos θ + sin θ)(−sinθ) + (cos θ)(− sin θ + cos θ)]
= 2√2V 2
g
[− sin θ cos θ − sin2 θ − sin θ cos θ + cos2 θ
]= 2√2V 2
g
[cos2 θ − sin2 θ − 2 sin θ cos θ
]
= 2√2V 2
g[cos 2θ − sin 2θ]
Paul Hancock (Woonona High School) Series T1 2017 23 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(iii) Show that dDdθ = 2
√2V 2
g (cos 2θ − sin 2θ).
This simply requires us to use the product rule on our answer to (ii).
dD
dθ= 2√2V 2
g[(cos θ + sin θ)(−sinθ) + (cos θ)(− sin θ + cos θ)]
= 2√2V 2
g
[− sin θ cos θ − sin2 θ − sin θ cos θ + cos2 θ
]= 2√2V 2
g
[cos2 θ − sin2 θ − 2 sin θ cos θ
]= 2√2V 2
g[cos 2θ − sin 2θ]
Paul Hancock (Woonona High School) Series T1 2017 23 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(iv) Show that D has a maximum value and find the value of θ for which thisoccurs.
From (iii) it should be clear that the stationary point(s) occur whencos 2θ = sin 2θ, ie. θ = π
8 .
Now we just need to find d2Ddθ2
(π8
)to show θ = π
8 produces a maximum value.
d2D
dθ2 = −4√2V 2
g[sin 2θ + cos 2θ]
d2D
dθ2
(π8
)= −4
√2V 2
g
[sin
π
4+ cos
π
4
]< 0
Paul Hancock (Woonona High School) Series T1 2017 24 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(iv) Show that D has a maximum value and find the value of θ for which thisoccurs.
From (iii) it should be clear that the stationary point(s) occur whencos 2θ = sin 2θ, ie. θ = π
8 .
Now we just need to find d2Ddθ2
(π8
)to show θ = π
8 produces a maximum value.
d2D
dθ2 = −4√2V 2
g[sin 2θ + cos 2θ]
d2D
dθ2
(π8
)= −4
√2V 2
g
[sin
π
4+ cos
π
4
]< 0
Paul Hancock (Woonona High School) Series T1 2017 24 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(iv) Show that D has a maximum value and find the value of θ for which thisoccurs.
From (iii) it should be clear that the stationary point(s) occur whencos 2θ = sin 2θ, ie. θ = π
8 .
Now we just need to find d2Ddθ2
(π8
)to show θ = π
8 produces a maximum value.
d2D
dθ2 = −4√2V 2
g[sin 2θ + cos 2θ]
d2D
dθ2
(π8
)= −4
√2V 2
g
[sin
π
4+ cos
π
4
]< 0
Paul Hancock (Woonona High School) Series T1 2017 24 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(iv) Show that D has a maximum value and find the value of θ for which thisoccurs.
From (iii) it should be clear that the stationary point(s) occur whencos 2θ = sin 2θ, ie. θ = π
8 .
Now we just need to find d2Ddθ2
(π8
)to show θ = π
8 produces a maximum value.
d2D
dθ2 = −4√2V 2
g[sin 2θ + cos 2θ]
d2D
dθ2
(π8
)= −4
√2V 2
g
[sin
π
4+ cos
π
4
]< 0
Paul Hancock (Woonona High School) Series T1 2017 24 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(iv) Show that D has a maximum value and find the value of θ for which thisoccurs.
From (iii) it should be clear that the stationary point(s) occur whencos 2θ = sin 2θ, ie. θ = π
8 .
Now we just need to find d2Ddθ2
(π8
)to show θ = π
8 produces a maximum value.
d2D
dθ2 = −4√2V 2
g[sin 2θ + cos 2θ]
d2D
dθ2
(π8
)= −4
√2V 2
g
[sin
π
4+ cos
π
4
]
< 0
Paul Hancock (Woonona High School) Series T1 2017 24 / 24
SHM and Projectile Motion Standard Questions
2014 - Question 14a
(iv) Show that D has a maximum value and find the value of θ for which thisoccurs.
From (iii) it should be clear that the stationary point(s) occur whencos 2θ = sin 2θ, ie. θ = π
8 .
Now we just need to find d2Ddθ2
(π8
)to show θ = π
8 produces a maximum value.
d2D
dθ2 = −4√2V 2
g[sin 2θ + cos 2θ]
d2D
dθ2
(π8
)= −4
√2V 2
g
[sin
π
4+ cos
π
4
]< 0
Paul Hancock (Woonona High School) Series T1 2017 24 / 24
SHM and Projectile Motion Finale
Good Luck
And in the words of Douglas Adams:
Paul Hancock (Woonona High School) Series T1 2017 25 / 24
SHM and Projectile Motion Finale
Good Luck
And in the words of Douglas Adams: "So long and thanks for all the fish."
Paul Hancock (Woonona High School) Series T1 2017 25 / 24
SHM and Projectile Motion Finale
Good Luck
And in the words of Douglas Adams: "So long and thanks for all the fish."
I mean: "Don’t panic."
Paul Hancock (Woonona High School) Series T1 2017 25 / 24