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

To find the acceleration when the particle is first at rest, we note that the particle is at rest when its velocity is zero: $v = \frac{dx}{dt} = 0.$
We know that:
$v = \frac{dx}{dt} = 2 imes 3 ext{ cos } 3t = 6 ext{ cos } 3t.$
Setting this to zero gives: $6 ext{ cos } 3t = 0 \Rightarrow ext{cos } 3t = 0.$
The first occurrence happens at: $3t = \frac{ ext{π}}{2} \Rightarrow t = \frac{ ext{π}}{6}.$
Next, we compute the acceleration using: $a = \frac{d^2x}{dt^2} = \frac{dv}{dt} = -6 imes 3 ext{ sin } 3t = -18 ext{ sin } 3t.$
Substituting $t = \frac{ ext{π}}{6}$ results in: $a = -18 ext{ sin } \left( 3 \cdot \frac{ ext{π}}{6} \right) = -18 ext{ sin } \left( \frac{ ext{π}}{2} \right) = -18.$
Therefore, the acceleration at the first point of rest is -18 m/s².

To find the volume of the solid formed by rotating the region bounded by $y = ext{cos } 4x$ and the $x$-axis from $x = 0$ to $x = \frac{ ext{π}}{2}$ around the $x$-axis, we can use the disk method:

The volume $V$ is given by: $V = \pi \int_0^{\frac{ ext{π}}{2}} (y^2) \, dx = \pi \int_0^{\frac{ ext{π}}{2}} \left(\text{cos } 4x\right)^2 \, dx.$
Using the identity $ext{cos}^2 \theta = \frac{1 + \text{cos}(2\theta)}{2}$, we rewrite: $V = \pi \int_0^{\frac{ ext{π}}{2}} \frac{1 + \text{cos} (8x)}{2} \, dx.$
Evaluating the integral, we get: $V = \frac{\pi}{2} \left[ x + \frac{\text{sin}(8x)}{8} \right]_0^{\frac{ ext{π}}{2}} = \frac{\pi}{2} \left[ \frac{ ext{π}}{2} + 0 \right] = \frac{\pi^2}{4}.$
Thus, the volume of the solid is: $V = \frac{\pi^2}{4} \, \text{ cubic units.}$

To express $v^2$ in terms of $x$, we begin with the equation of acceleration: $\frac{d^2x}{dt^2} = 2 - \frac{x}{2}.$
We rewrite this in terms of velocity as: $\frac{dv}{dt} = 2 - \frac{x}{2}.$
Using the chain rule: $\frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}.$
Setting the equations equal gives: $v \frac{dv}{dx} = 2 - \frac{x}{2}.$
Separating the variables and integrating, we get: $\int v \, dv = \int \left(2 - \frac{x}{2}\right) \, dx.$
This leads to: $\frac{1}{2} v^2 = 2x - \frac{x^2}{4} + C.$ Given that $v = 4$ when $x = 0$, we find: $C = \frac{1}{2} \cdot 4^2 = 8.$
Now substituting back:
$\frac{1}{2} v^2 = 2x - \frac{x^2}{4} + 8.$
Thereafter, multiplying through by 2 gives: $v^2 = 4x - \frac{x^2}{2} + 16.$
Thus, the final expression for $v^2$ in terms of $x$ is: $v^2 = 4x - \frac{x^2}{2} + 16.$