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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 when the particle first returns to the origin, we need to determine the time it takes for the displacement function, given by , to return to zero. The particle returns to the origin when .
[ 2 \text{ sin } 3t = 0 ] This occurs when [3t = n\pi,] for integer values of . The first return to the origin occurs at , which gives: [ t = \frac{\pi}{3} ] To calculate the total distance travelled, we will find the displacement from to : Since the particle reaches a maximum amplitude of 2 meters, the total distance is twice this value:
Total Distance = meters.
Step 2
What is the acceleration of the particle when it is first at rest?
Answer
The acceleration of the particle is the second derivative of the displacement with respect to time. First, we find the velocity by differentiating the displacement: [ v(t) = \frac{dx}{dt} = 2 \cdot 3 \text{ cos } 3t = 6 \text{ cos } 3t ] The particle is at rest when , which happens when: [ 6 \text{ cos } 3t = 0 ] This occurs at (when ). Now, we calculate the acceleration: [ a(t) = \frac{dv}{dt} = -18 \text{ sin } 3t ] Substituting into the acceleration equation: [ a\left(\frac{\pi}{6}\right) = -18 \text{ sin } \left(3 \cdot \frac{\pi}{6}\right) = -18 \text{ sin } \left(\frac{\pi}{2}\right) = -18 ] Thus, the acceleration at this point is -18 m/s².
Step 3
Find the volume of the solid.
Answer
The volume of the solid formed by rotating the area bounded by and the -axis from to is given by the formula: [ V = \pi \int_0^{\frac{\pi}{2}} (\text{cos } 4x)^2 , dx ] Using the double angle formula: [ (\text{cos } 4x)^2 = \frac{1 + \text{cos } 8x}{2} ] Thus, [ V = \pi \int_0^{\frac{\pi}{2}} \frac{1 + \text{cos } 8x}{2} , dx ] This simplifies to: [ V = \frac{\pi}{2} \left[ x + \frac{1}{8} \text{sin } 8x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2} \left[ \frac{\pi}{2} + 0 - (0 + 0) \right] = \frac{\pi^2}{4} ] Thus, the volume of the solid is cubic units.
Step 4
Given that $v = 4$ when $x = 0$, express $v^2$ in terms of $x$.
Answer
We start with the acceleration equation: [ \frac{d^{2}x}{dt^{2}} = 2 - \frac{x}{2}. ] We recognize that acceleration can also be expressed in terms of velocity: [ a = v \frac{dv}{dx}. ] Setting these expressions equal gives: [ v \frac{dv}{dx} = 2 - \frac{x}{2}. ] Rearranging terms, we get: [ v , dv = \left(2 - \frac{x}{2}\right) dx. ] Integrating both sides yields: [ \int v , dv = \int \left(2 - \frac{x}{2}\right) dx ] Integrating gives: [ \frac{1}{2} v^2 = 2x - \frac{x^2}{4} + C. ] Using the condition when , we substitute to find: [ \frac{1}{2} (4^2) = C, ; \Rightarrow C = 8. ] Thus, the relationship is: [ v^2 = 16 - 4x + 2x^2. ] We can conclude with expressed in terms of :