SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Mechanics, EXT2 M1 2024 HSC 13b

A particle is moving in simple harmonic motion, described by  \(\ddot{x}=-4(x+1)\).

When the particle passes through the origin, the speed of the particle is 4 m s\(^{-1}\).

What distance does the particle travel during a full period of its motion?   (3 marks)

Show Answers Only

\(4 \sqrt{5} \text{ units }\)

Show Worked Solution

  \(\ddot{x}\) \(=-4(x+1)\)
  \(\dfrac{d}{dx}\left(\dfrac{1}{2} v^2\right)\) \(=-4 x-4\)
  \(\dfrac{1}{2} v^2\) \(=\displaystyle \int-4 x-4\, d x\)
  \(\dfrac{1}{2} v^2\) \(=-2 x^2-4 x+c\)
  \(v^2\) \(=-4 x^2-8 x+2c\)

 
\(\text {When } x=0, \quad v=4\ \ \Rightarrow \ c=8\)

\(v^2=-4 x^2-8 x+16\)

\(\text {Find } x \text { when } v=0:\)

\(-4 x^2-8 x+16\) \(=0\)  
\(x^2+2 x-4\) \(=0\)  

 
\(\Rightarrow\ x=\dfrac{-2 \pm \sqrt{2^2+4 \cdot 1 \cdot 4}}{2}=-1 \pm \sqrt{5}\)

\(\Rightarrow\ \text{Amplitude}=\sqrt{5}\)

\(\therefore \text {Distance travelled in full period }=4 \sqrt{5} \text{ units }\)

Mechanics, EXT2* M1 2014 HSC 12a

 A particle is moving in simple harmonic motion about the origin, with displacement `x` metres. The displacement is given by  `x = 2 sin 3t`, where `t` is time in seconds. The motion starts when  `t = 0`.

  1. What is the total distance travelled by the particle when it first returns to the origin?   (1 mark)

  2. What is the acceleration of the particle when it is first at rest?    (2 marks)

Show Answers Only
  1. `4\ text(m)`
  2. `-18\ text(m/s²)`
Show Worked Solution
i.    `x = 2 sin 3t`

 
`text(At)\ \ t = 0,\ x = 0`

`text(Amplitude) = 2`

`:.\ text(Distance travelled)`

`= 2 xx text(amplitude)`

`= 4\ text(m)`

 

ii.    `x` `= 2 sin 3t`
  `v` `= 6 cos 3t` 
  `ddot x` `= -18 sin 3t`

 
`text(Particle first comes to rest when)\ v = 0`

`6 cos 3t` `= 0`
`cos 3t` `= 0`
`3t` `= pi/2`
`t` `= pi/6`

 
`text(When)\ \  t = pi/6`

`ddot x` `= -18 * sin (3 xx pi/6)`
  `= -18 sin (pi/2)`
  `= -18\ text(m/s²)`

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)`