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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 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 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 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`