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Mechanics, EXT2 M1 2025 HSC 4 MC

A particle in simple harmonic motion has speed \(v \ \text{ms}^{-1}\), given by  \(v^2=-x^2+2 x+8\)  where \(x\) is the displacement from the origin in metres.

What is the amplitude of the motion?

  1. 1 m
  2. 3 m
  3. 6 m
  4. 9 m
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\(B\)

Show Worked Solution

\(\text{Using} \ \ v^2=n^2\left(a^2-(x-c)^2\right):\)

\(v^2\) \(=-x^2+2 x+8\)
  \(=9-\left(x^2-2 x+1\right)\)
  \(=9-(x-1)^2\)

 
\(\therefore a^2 = 9\ \ \Rightarrow\ \ a=3\)

\(\Rightarrow B\)

Mechanics, EXT2 M1 2023 HSC 11e

A particle moves in simple harmonic motion described by the equation

\( \ddot{x}=-9(x-4) . \)

Find the period and the central point of motion.   (2 marks)

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\(\text{Period}\ = \dfrac{2 \pi}{3} \)

\(\text{Centre of Motion:}\ x=4 \) 

Show Worked Solution

\( \ddot{x}=-9(x-4) \)

\( \Rightarrow\ n=3,\ \ c=4 \)

\(\text{Period}\ = \dfrac{2 \pi}{3} \)

\(\text{Centre of Motion:}\ x=4 \) 

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)

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`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 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.
  3. What is the earliest time tomorrow at which the tide is increasing at the fastest rate?   (2 marks)
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a.    `text(See Worked Solutions)`

b.    `11:22:30\ text(am)`

Show Worked Solution

a.    `text(High tide = 9m,  Low tide = 1 m)`

`A = (9-1)/2 = 4\ text(m)`

`\text{Since}\ \ 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)`
 

b.    `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 when)\ \ sin ((4 pi)/25 t) = -1:`

`(4 pi)/25 t= (3 pi)/2\ \ =>\ \ t=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
  2.    `y = 7 + 3 cos\ ((4pit)/(25))`, where `t` is the number of hours after high tide.   (2 marks)

  3. 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)

  4. 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.
  5. 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)

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a.    `text{Proof (See Worked Solutions)}`

b.    `text{Proof (See Worked Solutions)}`

c.    `4:28\ text(am)`

Show Worked Solution

a.    `text(Period)`

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

`text{Amplitude}\ (a) = 1/2(10-4)= 3`

 
`y=10\ \ \text{at}\ \ t=0\ \text{(low tide):`

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

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

b.    
        Calculus in the Physical, EXT1 2004 HSC 7a Answer

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

`7 + 3cos((4pit)/25)` `= 8.5`
`3cos((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(Earliest time a ship can leave is 4:05 am  … as required.)`
 

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

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

`7 + 3cos((4pit)/(25))` `=6`
`3cos((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(Harbour entrance depth 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 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
  4. `v^2 = n^2(a^2-(x-c)^2)`  where `a`, `c` and `n` are positive constants.
  5. What are the values of `a`, `c` and `n`?   (3 marks)

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a.    `3\ text(or)\ 7`

b.    `V = sqrt11\ text(m/s)`

c.    `a = 2, c = 5, n = (sqrt11)/2`

Show Worked Solution

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

`:. x = 3\ \ text(or)\ \ 7`
 

b.    `text(Maximum speed occurs when)`

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

 

c.    `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)`
`11` `= 4n^2`
`n^2` `= 11/4`
`:.n`  `= sqrt11/2`

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)

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

b.    `1.5\ text(seconds)\ text{(1 d.p.)}`

Show Worked Solution

a.    `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)`

 

b.    `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.)}`