HSC Revision Day Simple Harmonic Motion and Projectile Motionweb/@eis/documents/... · SHM and Projectile Motion Review HSC Question Analysis...

SHM and Projectile Motion

HSC Revision DaySimple Harmonic Motion and Projectile Motion

Woonona High School

Teacher: Paul HancockSchool: Woonona High SchoolE-mail: paul.hancock5@det.nsw.edu.au

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.

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.

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

"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:

dt= dv

dxdxdt

= v dvdx

I dvdt

dt= dv

dxdxdt

= v dvdx

I dvdt

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

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

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

= v dvdx

I dvdt

dt= dv

dxdxdt

= v dvdx

I dvdt

dt= dv

dxdxdt

= v dvdx

I dvdt

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.

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.

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

Answer:

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

Answer:

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:

n = 2π4 = π

2 Answer:

n = 2π4 = π

2 Answer: (A)

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:

We know v must be of the form v = an cos(nt). So, v = 5π3 cos(π3 t)

Answer:

3 cos(π3 t)

Answer:

3 cos(π3 t)

Answer: (A)

SHM and Projectile Motion Standard Questions

2014 - Question 12a

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.

2012 - Question 12a

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.

2012 - Question 12a

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.

2012 - Question 12a

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.

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

2012 - Question 12a

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

2012 - Question 12a

2 sin 3t = 2sin 3t = 1

3t =π

2012 - Question 12a

2 sin 3t = 2

sin 3t = 1

3t =π

2012 - Question 12a

2 sin 3t = 2sin 3t = 1

3t =π

2012 - Question 12a

2 sin 3t = 2sin 3t = 1

3t =π

2012 - Question 12a

2 sin 3t = 2sin 3t = 1

3t =π

2013 - Question 12e

We can solve this problem in 2 ways.

v2 = −9(x2 − k

= −9

x2 −

hence, n = 3 and so T = 2π3

(12v2)

2− 9

= −9x

2013 - Question 12e

v2 = −9(x2 − k

= −9

x2 −

(12v2)

2− 9

= −9x

2013 - Question 12e

v2 = −9(x2 − k

= −9

x2 −

(12v2)

2− 9

= −9x

2013 - Question 12e

v2 = −9(x2 − k

= −9

x2 −

(12v2)

2− 9

= −9x

2013 - Question 12e

v2 = −9(x2 − k

= −9

x2 −

hence, n = 3 and so T = 2π3 .

(12v2)

2− 9

= −9x

2013 - Question 12e

v2 = −9(x2 − k

= −9

x2 −

(12v2)

2− 9

= −9x

2013 - Question 12e

v2 = −9(x2 − k

= −9

x2 −

(12v2)

2− 9

= −9x

2013 - Question 12e

v2 = −9(x2 − k

= −9

x2 −

(12v2)

2− 9

= −9x

2013 - Question 12e

v2 = −9(x2 − k

= −9

x2 −

(12v2)

2− 9

= −9x

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)

2012 - Question 13c

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)

2012 - Question 13c

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)

2012 - Question 13c

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)

2012 - Question 13c

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)

2012 - Question 13c

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)

2012 - Question 13c

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)

2012 - Question 13c

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)

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)

2012 - Question 13c

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)

2012 - Question 13c

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)

2012 - Question 13c

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)

2012 - Question 13c

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)

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

dt= u sinα− gt

u sinα− gt = 0

t =u sinα

2013 - Question 13c

y = ut sinα− g

dt= u sinα− gt

u sinα− gt = 0

t =u sinα

2013 - Question 13c

y = ut sinα− g

dt= u sinα− gt

u sinα− gt = 0

t =u sinα

2013 - Question 13c

y = ut sinα− g

dt= u sinα− gt

u sinα− gt = 0

t =u sinα

2013 - Question 13c

y = ut sinα− g

dt= u sinα− gt

u sinα− gt = 0

t =u sinα

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

Since the particles collide at maximum height, then those two times should be thesame.

u sinαg

=w sinβ

u sinα = w sinβ

2013 - Question 13c

u sinαg

=w sinβ

u sinα = w sinβ

2013 - Question 13c

u sinαg

=w sinβ

u sinα = w sinβ

2013 - Question 13c

u sinαg

=w sinβ

u sinα = w sinβ

2013 - Question 13c

u sinαg

=w sinβ

u sinα = w sinβ

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.

(u sinα

)cosα+ w

(w sinβ

)cosβ

=u sinα× u cosα

w sinβ × w cosβg

=w sinβ × u cosα

u sinα× w cosβg

g(sinβ × cosα+ sinα× cosβ)

gsin(α+ β)

g sin(α+ β).

(u sinα

)cosα+ w

(w sinβ

)cosβ

=u sinα× u cosα

w sinβ × w cosβg

=w sinβ × u cosα

u sinα× w cosβg

gsin(α+ β)

g sin(α+ β).

(u sinα

)cosα+ w

(w sinβ

)cosβ

=u sinα× u cosα

w sinβ × w cosβg

=w sinβ × u cosα

u sinα× w cosβg

gsin(α+ β)

g sin(α+ β).

(u sinα

)cosα+ w

(w sinβ

)cosβ

=u sinα× u cosα

w sinβ × w cosβg

=w sinβ × u cosα

u sinα× w cosβg

gsin(α+ β)

g sin(α+ β).

(u sinα

)cosα+ w

(w sinβ

)cosβ

=u sinα× u cosα

w sinβ × w cosβg

=w sinβ × u cosα

u sinα× w cosβg

gsin(α+ β)

g sin(α+ β).

(u sinα

)cosα+ w

(w sinβ

)cosβ

=u sinα× u cosα

w sinβ × w cosβg

=w sinβ × u cosα

u sinα× w cosβg

gsin(α+ β)

2014 - Question 14a

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 .

V cos θ

y = −12g( 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

2V 2 sec2 θ.

V cos θ

y = −12g( x

V cos θ

)sin θ

y = − gx2

2V 2 cos2 θ+ x

sin θcos θ

y = − gx2

2V 2 sec2 θ.

V cos θ

y = −12g( x

V cos θ

)sin θ

y = − gx2

2V 2 cos2 θ+ x

sin θcos θ

y = − gx2

2V 2 sec2 θ.

V cos θ

y = −12g( x

V cos θ

)sin θ

y = − gx2

2V 2 cos2 θ+ x

sin θcos θ

y = − gx2

2V 2 sec2 θ.

V cos θ

y = −12g( x

V cos θ

)sin θ

y = − gx2

2V 2 cos2 θ+ x

sin θcos θ

y = − gx2

2V 2 sec2 θ.

V cos θ

y = −12g( x

V cos θ

)sin θ

y = − gx2

2V 2 cos2 θ+ x

sin θcos θ

y = − gx2

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=

sec2θ+

tan θsec2 θ

(cos2 θ + cos θ sin θ

gcos θ(cos θ + sin θ)

√2V 2

x tan θ − gx2

2V 2 sec2 θx2 − (1+ tan θ)x = 0

x =2V 2(1+ tan θ)

g sec2 θ=

sec2θ+

tan θsec2 θ

√2V 2

x tan θ − gx2

2V 2 sec2 θ = −x

2V 2 sec2 θx2 − (1+ tan θ)x = 0

x =2V 2(1+ tan θ)

g sec2 θ=

sec2θ+

tan θsec2 θ

√2V 2

x tan θ − gx2

2V 2 sec2 θx2 − (1+ tan θ)x = 0

x =2V 2(1+ tan θ)

g sec2 θ=

sec2θ+

tan θsec2 θ

√2V 2

x tan θ − gx2

2V 2 sec2 θx2 − (1+ tan θ)x = 0

x =2V 2(1+ tan θ)

g sec2 θ=

sec2θ+

tan θsec2 θ

√2V 2

x tan θ − gx2

2V 2 sec2 θx2 − (1+ tan θ)x = 0

x =2V 2(1+ tan θ)

g sec2 θ

sec2θ+

tan θsec2 θ

√2V 2

x tan θ − gx2

2V 2 sec2 θx2 − (1+ tan θ)x = 0

x =2V 2(1+ tan θ)

g sec2 θ=

sec2θ+

tan θsec2 θ

√2V 2

x tan θ − gx2

2V 2 sec2 θx2 − (1+ tan θ)x = 0

x =2V 2(1+ tan θ)

g sec2 θ=

sec2θ+

tan θsec2 θ

x =2V 2

√2V 2

x tan θ − gx2

2V 2 sec2 θx2 − (1+ tan θ)x = 0

x =2V 2(1+ tan θ)

g sec2 θ=

sec2θ+

tan θsec2 θ

2014 - Question 14a

(ii) Show that D = 2√2V 2

D =√x2 + y2

x2 + (−x)2

=√2x2

=√2x

= 2√2V 2

2014 - Question 14a

D =√x2 + y2

=√x2 + (−x)2

=√2x2

=√2x

= 2√2V 2

2014 - Question 14a

D =√x2 + y2

=√x2 + (−x)2

=√2x2

=√2x

= 2√2V 2

2014 - Question 14a

D =√x2 + y2

=√x2 + (−x)2

=√2x2

=√2x

= 2√2V 2

2014 - Question 14a

D =√x2 + y2

=√x2 + (−x)2

=√2x2

=√2x

= 2√2V 2

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).

dθ= 2√2V 2

g[(cos θ + sin θ)(−sinθ) + (cos θ)(− sin θ + cos θ)]

= 2√2V 2

[− sin θ cos θ − sin2 θ − sin θ cos θ + cos2 θ

]= 2√2V 2

[cos2 θ − sin2 θ − 2 sin θ cos θ

]= 2√2V 2

g[cos 2θ − sin 2θ]

2014 - Question 14a

√2V 2

dθ= 2√2V 2

= 2√2V 2

]= 2√2V 2

2014 - Question 14a

√2V 2

dθ= 2√2V 2

= 2√2V 2

]= 2√2V 2

2014 - Question 14a

√2V 2

dθ= 2√2V 2

= 2√2V 2

]= 2√2V 2

2014 - Question 14a

√2V 2

dθ= 2√2V 2

= 2√2V 2

]= 2√2V 2

= 2√2V 2

2014 - Question 14a

√2V 2

dθ= 2√2V 2

= 2√2V 2

]= 2√2V 2

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

Now we just need to find d2Ddθ2

)to show θ = π

8 produces a maximum value.

dθ2 = −4√2V 2

g[sin 2θ + cos 2θ]

)= −4

√2V 2

4+ cos

2014 - Question 14a

)to show θ = π

dθ2 = −4√2V 2

)= −4

√2V 2

4+ cos

2014 - Question 14a

)to show θ = π

dθ2 = −4√2V 2

)= −4

√2V 2

4+ cos

2014 - Question 14a

)to show θ = π

dθ2 = −4√2V 2

)= −4

√2V 2

4+ cos

2014 - Question 14a

)to show θ = π

dθ2 = −4√2V 2

)= −4

√2V 2

4+ cos

2014 - Question 14a

)to show θ = π

dθ2 = −4√2V 2

)= −4

√2V 2

4+ cos

SHM and Projectile Motion Finale

Good Luck

And in the words of Douglas Adams:

Good Luck

And in the words of Douglas Adams: "So long and thanks for all the fish."

Good Luck

And in the words of Douglas Adams: "So long and thanks for all the fish."

I mean: "Don’t panic."

HSC Revision Day Simple Harmonic Motion and Projectile Motionweb/@eis/documents/... · SHM and Projectile Motion Review HSC Question Analysis...

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