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Mechanics, EXT2 M1 2025 HSC 15b

A particle moves in simple harmonic motion about the origin with amplitude \(A\), and it completes two cycles per second. When it is \(\dfrac{1}{4}\) metres from the origin, its speed is half its maximum speed.

Find the maximum positive acceleration of the particle during its motion.   (4 marks)

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\(a_{\text{max}}=\dfrac{8 \pi^2}{\sqrt{3}}\)

Show Worked Solution

\(v^2=-n^2\left(x^2-A^2\right)\)

\(\operatorname{Period}\ (T)=\dfrac{1}{2} \ \Rightarrow \ \dfrac{2 \pi}{n}=\dfrac{1}{2} \ \Rightarrow \ n=4 \pi\)

\(\text{Max velocity occurs at}\ \  x=0:\)

   \(v_{\text{max}}^2=-(4 \pi)^2\left(0-A^2\right)=16 \pi^2 A^2\)

   \(v_{\text{max}}=\sqrt{16 \pi^2 A^2}=4 \pi A\)
 

\(\text{At} \ \ x=\dfrac{1}{4}, \ v=\dfrac{1}{2} \times 4 \pi A=2 \pi A\)

\((2 \pi A)^2\) \(=-(4 \pi)^2\left(\dfrac{1}{16}-A^2\right)\)
\(\dfrac{A^2}{4}\) \(=A^2-\dfrac{1}{16}\)
\(\dfrac{3 A^2}{4}\) \(=\dfrac{1}{16}\)
\(A^2\) \(=\dfrac{1}{12}\)
\(A\) \(=\dfrac{1}{\sqrt{12}}\)

 

\(\text{Max positive acceleration occurs at} \ \ x=-\dfrac{1}{\sqrt{12}}:\)

\(a_{\text{max}}=-n^2 x=-(4 \pi)^2 \times-\dfrac{1}{\sqrt{12}}=\dfrac{8 \pi^2}{\sqrt{3}}\)

Mechanics, EXT2* M1 2019 HSC 12b

A particle is moving along the `x`-axis in simple harmonic motion. The position of the particle is given by

`x = sqrt 2 cos 3t + sqrt 6 sin 3t,` for  `t >= 0` 

  1. Write  `x`  in the form  `R cos(3t - alpha)`, where  `R > 0`  and  `0 < alpha < pi/2`.  (2 marks)

  2. Find the two values for  `x`  where the particle comes to rest.   (1 mark)

  3. When is the first time that the speed of the particle is equal to half of its maximum speed?  (2 marks)

Show Answers Only
  1. `x = 2 sqrt 2 cos (3t – pi/3)`
  2. `2 sqrt 2 or -2 sqrt 2`
  3. `t = pi/18`
Show Worked Solution

i.    `x = sqrt 2 cos 3t + sqrt 6 sin 3t`

`R cos (3t – alpha) = R cos alpha cos 3t + R sin alpha sin 3t`

`=> R cos alpha = sqrt 2`

`=> R sin alpha = sqrt 6`

`R^2 cos^2 alpha + R^2 sin^2 alpha` `= 2 + 6`
`R^2 (cos^2 alpha + sin^2 alpha)` `= 8`
`R` `=2sqrt2`

  

`2 sqrt 2 cos alpha` `= sqrt 2`
`cos alpha` `= 1/2`
`alpha` `= pi/3`

 
`:. x = 2 sqrt 2 cos (3t – pi/3)`

♦ Mean mark part (ii) 46%.

 

ii.    `text(At the extremities of the amplitude,)`
 

`text(the particle stops and reverses.)`

`:. v = 0\ \ text(when)\ \ x = 2 sqrt 2 or -2 sqrt 2`

 

iii.   `x = 2 sqrt 2 cos (3t – pi/3)`

♦ Mean mark part (iii) 49%.

`(dx)/(dt) = -6 sqrt 2 sin(3t – pi/3)`

 
`text(Max speed) = 6 sqrt 2`

`text(Find)\ \ t\ \ text(when)\ \ (dx)/(dt) = +-3 sqrt 2`

`-6 sqrt 2 sin (3t – pi/3)` `= 3 sqrt 2`
`sin(3t – pi/3)` `= -1/2`
`3t – pi/3` `= -pi/6`
`3t` `= pi/6`
`t` `= pi/18`

 

`-6 sqrt 2 sin (3t – pi/3)` `= -3 sqrt 2`
`sin(3t – pi/3)` `= 1/2`
`3t – pi/3` `= pi/6`
`3t` `= pi/2`
`t` `= pi/6`

 
`:. t = pi/18\ text{(1st time)}`

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)

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  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 2016 HSC 7 MC

The displacement `x` of a particle at time `t` is given by

`x = 5 sin 4t + 12 cos 4t`.

What is the maximum velocity of the particle?

  1. `13`
  2. `28`
  3. `52`
  4. `68`
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`C`

Show Worked Solution

`x = 5 sin 4t + 12 cos 4t`

`(dx)/(dt) = 20 cos 4t – 48 sin 4t`

 `=>\ text(Can be written in the form:)`

`A cos (4t + alpha),\ text(where)`

`A` `= sqrt (20^2 + 48^2)`
  `= 52`

 
`:. text(Max)\ v = 52\ text(ms)^-1`

`=>   C`

Mechanics, EXT2* M1 2005 HSC 5c

A particle moves in a straight line and its position at time  `t`  is given by

`x = 5 + sqrt 3 sin3t - cos 3t.`

  1. Express  `sqrt 3 sin3t − cos 3t`  in the form  `R sin(3t - alpha)`  where  `alpha`  is in radians.  (2 marks)

  2. The particle is undergoing simple harmonic motion. Find the amplitude and the centre of the motion.  (2 marks)

  3. When does the particle first reach its maximum speed after time  `t = 0`?  (1 mark)

Show Answers Only
  1. `2 sin ( 3t – pi/6)`
  2. `text(Amplitude) = 2\ text(units),\ \ \ text(Centre of the motion at)\ \ x = 5`
  3. `pi/18`
Show Worked Solution

i.    `x = 5 + sqrt 3 sin 3t – cos 3t`

`sqrt 3 sin 3t – cos 3t` `= R sin (3t – alpha)`
  `= R sin 3t cos alpha – R cos 3t sin alpha`

 
`=> R cos alpha = sqrt 3\ \ \ \ \ R sin alpha = 1`

`R^2 = sqrt 3^2 + 1^2 = 4`

`R = 2`

`=> 2 cos alpha` `= sqrt3`
`cos alpha` `= (sqrt3)/2`
`alpha` `= pi/6`

 
`:.\ 2 sin ( 3t – pi/6) = sqrt 3 sin 3t – cos 3t`

 

ii.  `x` `= 5 + sqrt 3 sin 3t – cos 3t`
  `= 5 + 2 sin (3t – pi/6)`

 
`text(Amplitude) = 2\ text(units)`

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

 

iii.  `text(Solution 1)`

`x = 5 + 2 sin (3t – pi/6)`

`dot x = 6 cos (3t – pi/6)`

 
`text(Maximum speed occurs when)`

`cos (3t – pi/6)` `= 1`
`3t – pi/6` `= 0`
`3t` `= pi/6`
`t` `= pi/18`

 
`text(Solution 2)`

`text(Maximum speed occurs at the)`

`text(centre of motion,)\ \ x = 5`

`5 + 2 sin (3t – pi/6)` `= 5`
`2 sin (3t – pi/6)` `= 0`
`sin (3t – pi/6)` `= 0`
`3t – pi/6` `= 0`
`t` `= pi/18`

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)

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  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 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)

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  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 2010 HSC 4a

A particle is moving in simple harmonic motion along the `x`-axis. 

Its velocity `v`, at `x`, is given by  `v^2 = 24 − 8x − 2x^2`. 

  1. Find all values of `x` for which the particle is at rest.   (1 mark)

  2. Find an expression for the acceleration of the particle, in terms of `x`.   (1 mark)

  3. Find the maximum speed of the particle.    (2 marks)

Show Answers Only
  1. `x = –6\ \ text(or)\ \ 2`
  2. `-4\ – 2x`
  3. `4 sqrt 2`
Show Worked Solution
(i)    `v^2 = 24\ – 8x\ – 2x^2`

`text(Find)\ \ x\ \ text(when)\ \ v=0`

`24 – 8x – 2x^2` `= 0`
`x^2 + 4x – 12` `= 0`
`(x + 6)(x – 2)` `= 0`
`x= -6\ \ text(or)\ \ 2`

 

`:.\ text(Particle at rest when)\ \ x = –6\ \ text(or)\ \ 2`

 

(ii)    `ddot x` `= d/(dx) (1/2 v^2)`
    `= d/(dx) (12 – 4x – x^2)`
    `= -4 – 2x`

 

(iii)   `text(Solution 1)`

`text(Max speed when)\ \ ddot x = 0`

`-4 – 2x` `= 0`
`2x` `= -4`
`x` `= -2`

`text(At)\ \ x = -2`

MARKER’S COMMENT: While most students found `x=-2` correctly, too many made errors substituting this back in or failed to finish the answer by taking the square root. BE CAREFUL! .
`v^2` `= 24 – 8(–2) – 2(–2)^2`
  `= 24 + 16 – 8`
  `= 32`
`v` `= +- sqrt32`
  `= +- 4 sqrt2`

`:.\ text(Maximum speed is)\ \ 4 sqrt2`

 

`text(Alternate Answer)`

`text(Maximum speed occurs when)`

`x = (-6+2)/2 = –2`

`text(At)\ \ x = –2`

`v^2` `= 32\ \ \ text{(see working above)}`
`v` `= +- 4 sqrt 2`

 

`:.\ text(Maximum speed is)\ 4sqrt2`