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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)

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

Show Worked Solution

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

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

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

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

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

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

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

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

Filed Under: Simple Harmonic Motion, Simple Harmonic Motion Tagged With: Band 4, Period and Cetre, smc-1059-20-Amplitude/Period, smc-1059-50-Distance Travelled, smc-7438-20-Amplitude/Period, smc-7438-50-Distance Travelled

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)

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

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a.    `4\ text(m)`

b.    `-18\ text(m/s²)`

Show Worked Solution
a.     `x = 2 sin 3t`

 
`text(At)\ \ t = 0,\ x = 0`

`text(Amplitude) = 2`

`:.\ text(Distance travelled)`

`= 2 xx text(amplitude)`

`= 4\ text(m)`

 

b.     `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)`

Filed Under: Simple Harmonic Motion, Simple Harmonic Motion, Simple Harmonic Motion EXT1 Tagged With: Band 4, smc-1059-40-At Rest/Endpoints, smc-1059-50-Distance Travelled, smc-7438-40-At Rest/Endpoints, smc-7438-50-Distance Travelled

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)

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  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)

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

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

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

b.    `A = 0,\ B = 2`

c.    `t = pi/(2n)`

d.    `text(8 units)`

Show Worked Solution
a.   `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)`

 

b.    `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%
 

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

 

d.    `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.

`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)`

Filed Under: Simple Harmonic Motion, Simple Harmonic Motion, Simple Harmonic Motion EXT1 Tagged With: Band 4, Band 5, Band 6, smc-1059-10-Prove/Identify SHM, smc-1059-50-Distance Travelled, smc-7438-10-Prove/Identify SHM, smc-7438-50-Distance Travelled

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