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A particle is moving in simple harmonic motion about the origin, with displacement $x$ metres - HSC - SSCE Mathematics Extension 1 - Question 12 - 2014 - Paper 1
Question 12
A particle is moving in simple harmonic motion about the origin, with displacement $x$ metres. The displacement is given by $x = 2 ext{ sin } 3t$, where $t$ is time... show full transcript
Worked Solution & Example Answer:A particle is moving in simple harmonic motion about the origin, with displacement $x$ metres - HSC - SSCE Mathematics Extension 1 - Question 12 - 2014 - Paper 1
Step 1
What is the total distance travelled by the particle when it first returns to the origin?
Answer
To find the total distance travelled by the particle when it first returns to the origin, we need to determine when the displacement function equals zero again after . The particle first returns to the origin when the sine function completes a full cycle. This occurs at the first positive root of the equation:
[ 3t = ext{π} ]
So,
[ t = \frac{ ext{π}}{3} ]
To find the total distance travelled, we calculate the distance from the origin to the peak at and return to the origin:
- From to , the particle moves upward, covering a distance of 2.
- From to , it returns to the origin, covering the same distance again.
Thus, the total distance travelled is:
[ 2 + 2 = 4 ext{ metres} ]
Step 2
What is the acceleration of the particle when it is first at rest?
Answer
The particle is at rest when its velocity is zero. The velocity can be found by differentiating the displacement function:
[ v(t) = \frac{dx}{dt} = 2 imes 3 \cos(3t) = 6 \cos(3t) ]
Setting gives:
[ \cos(3t) = 0 ]
This occurs when:
[
3t = \frac{\text{π}}{2} ext{ (first instance)}
]
So:
[ t = \frac{\text{π}}{6} ]
Now, we differentiate the velocity function to find the acceleration:
[ a(t) = \frac{dv}{dt} = -6 imes 3 \sin(3t) = -18 \sin(3t) ]
Substituting :
[ a\left(\frac{\text{π}}{6}\right) = -18 \sin\left(\text{π}/2\right) = -18 ext{ m/s}^2 ]
Step 3
Find the volume of the solid.
Answer
To find the volume of the solid formed by rotating the region bounded by and the x-axis from to about the x-axis, we use the formula for volume:
[ V = \pi \int_0^{\frac{\text{π}}{2}} (\cos(4x))^2 dx ]
Using the double angle formula: , we have:
[ V = \pi \int_0^{\frac{\text{π}}{2}} \frac{1 + \cos(8x)}{2} dx ]
This simplifies to:
[ V = \frac{\pi}{2} \left[ x + \frac{\sin(8x)}{8} \right]_0^{\frac{\text{π}}{2}} ]
Substituting the limits:
[ V = \frac{\pi}{2} \left[ \frac{\text{π}}{2} + 0 - (0 + 0)\right] = \frac{\text{π}^2}{4} ext{ cubic units} ]
Step 4
Given that v = 4 when x = 0, express v² in terms of x.
Answer
Starting from the given acceleration equation:
[ \frac{dv}{dt} = 2 - \frac{x}{2} ]
Using the chain rule for acceleration:
[ a = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v ]
Substituting and rearranging:
[
v \frac{dv}{dx} = 2 - \frac{x}{2}\
]
This leads us to integrate:
[ \int v , dv = \int \left(2 - \frac{x}{2}\right) dx ]
Thus,
[ v^2 = 2x - \frac{x^2}{4} + C ]
Applying the condition when , gives:
[ 16 = 0 + C \Rightarrow C = 16 ]
Therefore:
[ v^2 = 2x - \frac{x^2}{4} + 16\text{ m}^2/s^2 ]