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Mechanics, EXT2 M1 2023 HSC 6 MC

Which of the following functions does NOT describe simple harmonic motion?

  1. \(x=\cos ^2 t-\sin 2 t\)
  2. \(x=\sin 4 t+4 \cos 2 t\)
  3. \(x=2 \sin 3 t-4 \cos 3 t+5\)
  4. \(x=4 \cos \left(2 t+\dfrac{\pi}{2}\right)+5 \sin \left(2 t-\dfrac{\pi}{4}\right)\)
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\(B\)

Show Worked Solution

\(\text{By trial and error}\)

\(\text{Option}\ A:\)

\(x=\cos^2t-\sin\,2t=\dfrac{1}{2}\cos\,2t+\dfrac{1}{2}-\sin\,2t \)

\(\dot x=-\sin\,2t-2\cos\,2t \)

♦♦ Mean mark 34%.
\(\ddot x\) \(=-2\cos\,2t+4\sin\,2t\)  
  \(=-4\Big(\dfrac{1}{2}\cos\,2t-\sin\,2t\Big) \)  
  \(=-4\Big(x-\dfrac{1}{2}\Big) \ \ \ \text{(SHM)}\)  

 
\(\text{Similarly, options}\ C\ \text{and}\ D\ \text{can be differentiated to show} \)

\(\ddot x=-n^2(x-c) \)
 

\(\text{Consider option}\ B:\)

\(x=\sin\,4t+4\cos\,2t\)

\(\dot x=4\cos\,4t-8\sin\,2t\)

\(\ddot x\) \(=-16\sin\,4t-16\cos\,2t\)  
  \(= -16(\sin\,4t-\cos\,2t)\ \ \ \ \text{(not SHM)} \)  

 
\(\Rightarrow B\)

Mechanics, EXT2 M1 2020 HSC 13a

A particle is undergoing simple harmonic motion with period `frac{pi}{3}`. The central point of motion of the particle is at  `x = sqrt(3)`. When  `t = 0`  the particle has its maximum displacement of  `2 sqrt(3)`  from the central point of motion.

Find an equation for the displacement, `x`, of the particle in terms of `t`.    (3 marks)

Show Answers Only

`x = 2 sqrt(3) cos (6t) + sqrt(3)`

Show Worked Solution
`text{Period}` `= frac{pi}{3}`
`frac{2 pi}{n}` `= frac{pi}{3}`
`n` `= 6`

 
`text{Amplitude} = 2 sqrt(3)`

`text{Centre of motion} = sqrt(3)`

`text{S} text{ince maximum displacement at}\ \ t = 0,`

`x = 2 sqrt(3) cos (6t) + sqrt(3)`

Mechanics, EXT2* M1 2018 HSC 10 MC

A particle is moving in simple harmonic motion. The displacement of the particle is  `x`  and its velocity, `v`, is given by the equation  `v^2 = n^2 (2kx - x^2)`, where `n` and `k` are constants.

The particle is initially at  `x = k`.

Which function, in terms of time `t`, could represent the motion of the particle?

A.     `x = k cos (nt)`

B.     `x = k sin (nt) + k`

C.     `x = 2k cos (nt) - k`

D.     `x = 2k sin (nt) + k`

Show Answers Only

`B`

Show Worked Solution

`text(Completing the square):`

♦♦ Mean mark 33%.

`v^2 = n^2(2kx-x^2)`

    `=n^2(k^2-(x^2-2kx+k^2))`

    `=n^2(k^2-(x-k)^2)`
 

`:.\ text(The centre of motion is)\ \ x = k,\ text(amplitude) = k,`

`⇒  B`

Mechanics, EXT2* M1 2016 HSC 13a

The tide can be modelled using simple harmonic motion.

At a particular location, the high tide is 9 metres and the low tide is 1 metre.

At this location the tide completes 2 full periods every 25 hours.

Let `t` be the time in hours after the first high tide today.

  1. Explain why the tide can be modelled by the function  `x = 5 + 4cos ((4pi)/25 t)`.  (2 marks)

  2. The first high tide tomorrow is at 2 am.

     

    What is the earliest time tomorrow at which the tide is increasing at the fastest rate?  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `11:22:30\ text(am)`
Show Worked Solution
i.    `text(High tide)` `= 9\ text(m)`
  `text(Low tide)` `= 1\ text(m)`

 
`:. A = (9 – 1)/2 = 4\ text(m)`

`T = 25/2`

`:. (2 pi)/n` `= 25/2`
`n` `= (4 pi)/25`

 

`text(Centre of motion) = 5`

 

`text(S) text(ince high tide occurs at)\ \ t = 0,`

`x` `= 5 + 4 cos (nt)`
  `= 5 + 4 cos ((4 pi)/25 t)`

 

ii.   `x = 5 + 4 cos ((4 pi)/25 t)`

`(dx)/(dt)` `= -4 · (4 pi)/25 *sin ((4 pi)/25 t)`
  `= -(16 pi)/25 *sin ((4 pi)/25 t)`

 

`text(Tide increases at maximum rate)`

`text(when)\ \ sin ((4 pi)/25 t) = -1,`

`:. (4 pi)/25 t` `= (3 pi)/2`
  `= 75/8`
  `= 9\ text(hours 22.5 minutes)`

 

`:.\ text(Earliest time is)\ 11:22:30\ text(am)`

Mechanics, EXT2* M1 2004 HSC 7a

The rise and fall of the tide is assumed to be simple harmonic, with the time between successive high tides being 12.5 hours. A ship is to sail from a wharf to the harbour entrance and then out to sea. On the morning the ship is to sail, high tide at the wharf occurs at 2 am. The water depths at the wharf at high tide and low tide are 10 metres and 4 metres respectively.

  1. Show that the water depth, `y` metres, at the wharf is given by
     
         `y = 7 + 3 cos\ ((4pit)/(25))`, where  `t`  is the number of hours after high tide.  (2 marks)  

  2. An overhead power cable obstructs the ship’s exit from the wharf. The ship can only leave if the water depth at the wharf is 8.5 metres or less.

     

    Show that the earliest possible time that the ship can leave the wharf is 4:05 am.  (2 marks)

  3. At the harbour entrance, the difference between the water level at high tide and low tide is also 6 metres. However, tides at the harbour entrance occur 1 hour earlier than at the wharf. In order for the ship to be able to sail through the shallow harbour entrance, the water level must be at least 2 metres above the low tide level.

     

    The ship takes 20 minutes to sail from the wharf to the harbour entrance and it must be out to sea by 7 am. What is the latest time the ship can leave the wharf?  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `4:28\ text(am)`
Show Worked Solution

i.   `text(Period)`

`(2pi)/n =` ` 12.5`
`12.5n =` ` 2pi`
`25n =` ` 4pi`
`n =` ` (4pi)/25`

 
`text(Amplitude)`

`a` `= 1/2(10 − 4)`
  `= 3`

 
`=>\ text(Motion centres around)\ \ x = 7\ \ text(with)`

`y = 10\ \ text(when)\ \ t= 0`

 

`:.\ text(Water depth is given by)`

`y` `= 7+a cos nt`
  `= 7 + 3 cos\ ((4pit)/(25))\ \ …\ text(as required.)`

  

ii.  

 Calculus in the Physical, EXT1 2004 HSC 7a Answer

`text(Find)\ \ t\ \ text(when)\ \ y = 8.5:`

`8.5` `= 7 + 3\ cos\ ((4pit)/25)`
`3\ cos\ ((4pit)/25)` `= 1.5`
`cos\ ((4pit)/25)` `= 1/2`
`(4pit)/25` `=pi/3`
`t` `=(25pi)/(3 xx 4pi)`
  `=2 1/12`
  `= 2\ text(hrs 5 mins)`

 

`:.\ text(The earliest time the ship can leave)`

`text(is 4:05 am  … as required.)`

 

iii.  `text(2 metres above low tide = 6 m)`

`text(Find)\ t\ text(when)\ y = 6:`

`6 = 7 + 3\ cos\ ((4pit)/(25))`

`3\ cos\ ((4pit)/(25))` `= -1`
`cos\ ((4pit)/(25))` `= -1/3`
`(4pit)/25` `= 1.9106…`
`:.t` `= (25 xx 1.9106…)/(4pi)`
  `= 3.801…`
  `= 3\ text{hr 48 min  (nearest min)}`

 

`:.\ text(At the harbour entrance, water depth)`

`text(of 6 m occurs when)\ \ t = 2\ text(hr 48 min.)`

 

`:.\ text(Given 20 mins sailing time, the latest the ship can)`

`text(leave the wharf is at)\ \ t = 2\ text(hr 28 min, or 4:28 am.)`

Mechanics, EXT2* M1 2007 HSC 6a

A particle moves in a straight line. Its displacement, `x` metres, after `t` seconds is given by

`x = sqrt3\ sin\ 2t − cos\ 2t + 3`.

  1. Prove that the particle is moving in simple harmonic motion about  `x = 3`  by showing that   `ddot x = -4(x − 3)`.  (2 marks)

  2. What is the period of the motion?  (1 mark)

  3. Express the velocity of the particle in the form  `dotx = A\ cos\ (2t − α)`, where  `α`  is in radians.  (2 marks)

  4. Hence, or otherwise, find all times within the first  `pi`  seconds when the particle is moving at `2` metres per second in either direction.  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions.)}`
  2. `pi\ text(seconds)`
  3. `dot x = 4\ cos\ (2t − pi/6)`
  4. `t = pi/4, (5pi)/12, (3pi)/4, (11pi)/12\ text(seconds.)`
Show Worked Solution

i.   `text(Show)\ \ ddot x = -4(x − 3)`

`x` `= sqrt3\ sin\ 2t − cos\ 2t + 3`
`dot x` `= 2sqrt3\ cos\ 2t + 2\ sin\ 2t`
`ddot x` `= -4sqrt3\ sin\2t + 4\ cos\ 2t`
  `= -4(sqrt3\ sin\ 2t − cos\ 2t)`
  `= -4(sqrt3\ sin\ 2t − cos\ 2t + 3 − 3)`
  `= -4(x − 3)\ \ …text(as required)`

 

ii.  `text(Period)\ = (2pi)/n`

`n^2 = 4 ⇒ n = 2\ \ text{(part (i))}`

`:.\ text(Period)` `= (2pi)/2`
  `= pi\ \ text(seconds)`

 

iii.  `text(Write)\ \ dot x = 2sqrt3\ cos\ 2t + 2\ sin\ 2t`

`text(in form)\ \ \ A\ cos\ (2t − α)`

`A(cos\ 2t\ cos\ α + sin\ 2t\ sin\ α)` `= 2sqrt3\ cos\ 2t + 2\ sin\ 2t`
`cos\ 2t\ cos\ α + sin\ 2t\ sin\ α` `= (2sqrt3)/A\ cos\ 2t + 2/A\ sin\ 2t`
`⇒ cos\ α` `= (2sqrt3)/A`
`⇒ sin\ α` `= 2/A`
`((2sqrt3)/A)^2 + (2/A)^2` `= 1`
`(2sqrt3)^2 + 2^2` `= A^2`
`:.A` `= sqrt16`
  `= 4`

 

`:.cos\ α` `= (2sqrt3)/4 = sqrt3/2`
`α` `= pi/6`

`:. dot x = 4\ cos\ (2t − pi/6)`

 

(iv)  `text(Find)\ \ t\ \ text(when)\ \ dot x = ±2`

`text(If)\ \ dot x = 2`

`4\ cos\ (2t − pi/6)` `= 2`
`cos\ (2t − pi/6)` `= 1/2`
`2t − pi/6` `= pi/3, 2pi − pi/3`
`2t` `= pi/2, (11pi)/6`
`t` `= pi/4, (11pi)/12`

 

`text(If)\ \ dot x = -2`

`cos\ (2t − pi/6)` `= – 1/2`
`2t − pi/6` `= (2pi)/3, (4pi)/3`
`2t` `= (5pi)/6, (3pi)/2`
`t` `= (5pi)/12, (3pi)/4`

 

`:.t = pi/4, (5pi)/12, (3pi)/4, (11pi)/12\ \ text(seconds.)`

Mechanics, EXT2* M1 2015 HSC 13a

A particle is moving along the `x`-axis in simple harmonic motion. The displacement of the particle is `x` metres and its velocity is `v` ms`\ ^(–1)`. The parabola below shows `v^2` as a function of `x`.

2015 13a

  1. For what value(s) of `x` is the particle at rest?  (1 mark)

  2. What is the maximum speed of the particle?  (1 mark)

  3. The velocity `v` of the particle is given by the equation

         `v^2 = n^2(a^2 − (x −c)^2)`  where `a`, `c` and `n` are positive constants.

     

    What are the values of `a`, `c` and `n`?  (3 marks)

Show Answers Only
  1. `3\ text(or)\ 7`
  2. `V = sqrt11\ text(m/s)`
  3. `a = 2, c = 5,`
  4. `n = (sqrt11)/2`
Show Worked Solution

i.   `text(Particle is at rest when)\ v^2 = 0`

`:. x = 3\ \ text(or)\ \ 7`

 

ii.   `text(Maximum speed occurs when)`

`v^2` `= 11`
`v` `= sqrt11\ text(m/s)`

 

iii.  `v^2 = n^2(a^2 − (x − c)^2)`

♦ Mean mark 41%.

`text(Amplitude) = 2`

`:.a = 2`

`text(Centre of motion when)\ x = 5`

`:.c = 5`

`text(S)text(ince)\ \ v^2 = 11\ \ text(when)\ \ x = 5`

`11` `= n^2(2^2 − (5 − 5)^2)`
  `= 4n^2`
`n^2` `= 11/4`
`:.n`  `= sqrt11/2`

Mechanics, EXT2* M1 2014 HSC 7 MC

A particle is moving in simple harmonic motion with period 6 and amplitude 5.

Which is a possible expression for the velocity, `v`, of the particle?

  1. `v = (5pi)/3 cos (pi/3 t)`
  2. `v = 5 cos (pi/3 t)`
  3. `v = (5pi)/6 cos (pi/6 t)`
  4. `v = 5 cos (pi/6 t)`
Show Answers Only

`A`

Show Worked Solution

`text(By Elimination:)`

`text(General form is)\ \x = a sin (nt)`

♦ Mean mark 43%
COMMENT: Elimination can be a very effective and time efficient strategy for solving MC questions.

`text(Period) = 6\ text{(given)}`

`=> (2pi)/n` `= 6`
`n` `= pi/3`

 
`:.\ text(Cannot be)\ C\ text(or)\ D`

 

`text(Using)\ \ x = int v\ dt`

`text(Consider answer)\ B`

`x` `= 5 xx 3/pi sin (pi/3 t) + c`
  `text(Amplitude) = 15/pi`

 
`:.\ text(Cannot be)\ B`

 

`text(Consider answer)\ A`

`x` `= (5pi)/3 xx 3/pi sin (pi/3 t) + c`
  `= 5 sin (pi/3 t) + c`
 `:.\ text(Amplitude) = 5`

 
`=>  A`

Mechanics, EXT2* M1 2009 HSC 5a

The equation of motion for a particle moving in simple harmonic motion is given by

`(d^2x)/(dt^2) = -n^2x`

where  `n`  is a positive constant,  `x`  is the displacement of the particle and  `t`  is time.  

  1. Show that the square of the velocity of the particle is given by
     
         `v^2 = n^2 (a^2\ - x^2)`

     

    where  `v = (dx)/(dt)`  and  `a`  is the amplitude of the motion.   (3 marks)

  2. Find the maximum speed of the particle.     (1 mark)

  3. Find the maximum acceleration of the particle.    (1 mark)

  4. The particle is initially at the origin. Write down a formula for  `x`  as a function of  `t`, and hence find the first time that the particle’s speed is half its maximum speed.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `na`
  3. `n^2 a`
  4. `pi/(3n)`
Show Worked Solution
i.    `text(Show)\ \ v^2 = n^2 (a^2\ – x^2)`

`(d^2x)/(dt^2) = -n^2x`

`d/(dx) (1/2 v^2)` `= -n^2 x`
`1/2 v^2` `= int -n^2x\ dx`
  `= (-n^2 x^2)/2 + c`
`v^2` `= -n^2 x^2 + c`

 
`text(When)\ \ v = 0,\ \ x = a`

`0` `= -n^2a + c`
`c` `= n^2 a^2`
`:.\ v^2` `= -n^2 x^2 + n^2 a^2`
  `= n^2 (a^2\ – x^2)\ \ text(… as required)`

 

ii.    `text(Max speed when)\ \ x = 0`
`v^2` `= n^2 (a^2\ – 0)`
  `= n^2 a^2`
`:.v_text(max)` `= na`

 

iii.   `(d^2x)/(dt^2)\ \ text(is maximum at limits)\ \ (x = +-a)`
`(d^2x)/(dt^2)` `= |n^2 (a)|`
  `= n^2 a`

 

iv.   `x` `= a sin nt`
  `dot x` `= an cos nt`

 

`text(Find)\ \ t\ \ text(when)\ \ dot x = (na)/2`

`(na)/2` `= an cos nt`
`cos nt` `= 1/2`
`nt` `= pi/3`
`:.t` `= pi/(3n)`

Mechanics, EXT2* M1 2013 HSC 12e

A particle moves along a straight line. The displacement of the particle from the origin is `x`, and its velocity is `v`. The particle is moving so that  `v^2 + 9x^2 = k`, where `k` is a constant.

Show that the particle moves in simple harmonic motion with period  `(2pi)/3`.   (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution
`v^2 + 9x^2` `= k`
`v^2` `= k\ – 9x^2`
`1/2 v^2` `= 1/2k\ – 9/2 x^2`

 

`text(For SHM,)\ \ ddot x = -n^2x`

`ddot x` `= d/(dx) (1/2v^2)`
  `= -9x`
  `= -3^2 x \ \ \ text(… as required)`

 

`text(Period)\ (T)\ text(of SHM) = (2pi)/n`

`text(Here,)\ \ n=3`

`:.T= (2pi)/3\ \ \ text(… as required)`

Mechanics, EXT2* M1 2011 HSC 3a

The equation of motion for a particle undergoing simple harmonic motion is 

 `(d^2x)/(dt^2) = -n^2 x`,

where `x` is the displacement of the particle from the origin at time `t`, and `n` is a positive constant.

  1. Verify that  `x = A cos nt + B sin nt`, where `A` and `B` are constants, is a solution of the equation of motion.    (1 mark)

  2. The particle is initially at the origin and moving with velocity `2n`. 

     

    Find the values of `A` and `B` in the solution  `x = A cos nt + B sin nt`.    (2 marks)

  3. When is the particle first at its greatest distance from the origin?   (1 mark)

  4. What is the total distance the particle travels between  `t = 0`  and  `t = (2pi)/n`?   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `A = 0,\ B = 2`
  3. `t = pi/(2n)`
  4. `text(8 units)`
Show Worked Solution
i.   `x` `= A cos nt + B sin nt`
  `(dx)/(dt)` `= – An sin nt + Bn cos nt`
  `(d^2x)/(dt^2)` `= – An^2 cos nt\ – Bn^2 sin nt`
    `= -n^2 (A cos nt + B sin nt)`
    `= -n^2 x\ \ \ text(… as required)`

 

ii.   `text(At)\ \ t=0, \ x=0, \ v=2n:`

`x` `= Acosnt + Bsinnt`
`0` `= A cos 0 + B sin 0`
`:.A` `= 0`

  
`text(Using)\ \ (dx)/(dt) = Bn cos nt`

`2n` `= Bn cos 0`
`Bn` `= 2n`
`:.B` `= 2`
♦♦ Mean mark part (iii) 47%
 

iii.   `text(Max distance from origin when)\ (dx)/(dt) = 0`

`(dx)/(dt)` `= 2n cos nt`
`0` `= 2n cos nt`
`cos nt` `= 0`
`nt` `= pi/2,\ (3pi)/2,\ (5pi)/2`
`t` `= pi/(2n),\ (3pi)/(2n), …`

 

`:.\ text(Particle is first at greatest distance)`

`text(from)\ O\ text(when)\ t = pi/(2n).`

 

iv.  `text(Solution 1)`

`text(Find the distance travelled from)\ \ t=0\ \→\ \ t=(2pi)/n`

`text{(i.e. 1 full period)}`

♦♦ Mean mark 22%
MARKER’S COMMENT: Many students found the displacement at `t` rather than the distance travelled while another common error found distance as 2 x amplitude.

`text(S)text(ince)\ \ x=2 sin (nt)`

`=> text(Amplitude)=2`

`:.\ text(Distance travelled)=4 xx2=8\ text(units)`

 

`text(Solution 2)`

`text(At)\ t = 0,\ x = 0`

`text(At)\ t` `= pi/(2n)`
`x` `= 2 sin (n xx pi/(2n)) = 2`
`text(At)\ t` `= (3pi)/(2n)\ \ \ text{(i.e. the next time}\ \ (dx)/(dt) = 0 text{)}`
`x` `= 2 sin (n xx (3pi)/(2n)) = -2`
`text(At)\ t` `= (2pi)/n`
`x` `= 2 sin (n xx (2pi)/n) = 0`

 

`:.\ text(Total distance travelled)`

`= 2 + 4 + 2`
`= 8\ \ text(units)`

Mechanics, EXT2* M1 2012 HSC 13c

A particle is moving in a straight line according to the equation

`x = 5 + 6 cos 2t + 8 sin 2t`, 

where `x` is the displacement in metres and `t` is the time in seconds.

  1. Prove that the particle is moving in simple harmonic motion by showing that `x`  satisfies an equation of the form  `ddot x = -n^2 (x\ - c)`.  (2 marks)

  2. When is the displacement of the particle zero for the first time?    (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1.5\ text(seconds)\ text{(1 d.p.)}`
Show Worked Solution

i.  `text(Prove)\ ddot x = -n^2(x\ – c)`

`x` `= 5 + 6 cos 2t + 8 sin 2t`
`dot x` `= -12 sin 2t + 16 cos 2t`
`ddot x` `= – 24 cos 2t\ – 32 sin 2t`
  `= -4 (6 cos 2t + 8 sin 2t)`
  `= -2^2 (5 + 6 cos 2t + 8 sin 2t\ – 5)`
  `= -2^2 (x\ – 5)\ \ \ text(… as required)`

 

ii.  `text(Find)\ \ t\ \ text(when)\ \ x=0\ \ text(for 1st time:)`

♦ Mean mark 42%.
IMPORTANT: The critical insight required to solve `x=0` is to realise that the cosine of the difference between 2 angles, i.e. `cos (2t- theta)`, applies.
`5 + 6 cos 2t + 8 sin 2t` `= 0`
`6 cos 2t + 8 sin 2t` `= -5`
`6/10 cos 2t+ 8/10 sin 2t` `=-1/2`

 

 Calculus in the Physical World, EXT1 2012 HSC 13c Answer

`=>cos theta=6/10\ \ text(and)\ \ sin theta=8/10`
`cos 2t cos theta+sin 2t sin theta` `=- 1/2`
`cos(2t\ – theta)` `= – 1/2`

`text(S)text(ince)\ \ cos\ pi/3 = 1/2\ \ text(and)\ cos\ text(is negative)`

`text(in the 2nd and 3rd quadrants,)`

`=>2t\ – theta = pi\ – pi/3,\ pi + pi/3`

 

`text(We need the 1st time)\ \ x = 0`

`text(S)text(ince)\ \ cos theta` `= 6/10`
`theta` `= 0.9273…`

 

`:.\ 2t\ – 0.9273…` `= (2pi)/3`
`2t` `= (2pi)/3 + 0.9273…`
`t` `= 1/2 ((2pi)/3 + 0.9273…)`
  `= 1.5108…`
  `= 1.5\ text(seconds)\ text{(1 d.p.)}`