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

The particle is at rest when its velocity is 0. The velocity can be found by differentiating the displacement function:

$v(t) = \frac{dx}{dt} = 2 \cdot 3 \cos(3t) = 6 \cos(3t)$

Setting this equal to 0 gives:

$6 \cos(3t) = 0 \Rightarrow \cos(3t) = 0$

The first instance occurs when:

$3t = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{6}$

Then we calculate the acceleration, which is the second derivative of displacement:

$\ddot{x}(t) = \frac{d^2x}{dt^2} = -2 \cdot 3^2 \sin(3t) = -18 \sin(3t)$

Substituting $t = \frac{\pi}{6}$ gives:

$\ddot{x} = -18 \sin\left(\frac{\pi}{2}\right) = -18$

Therefore, the acceleration when the particle is first at rest is:

$-18 \text{ m/s}^2$.

To find the volume of the solid formed by rotating the region bounded by $y = \cos 4x$ and the $x$-axis about the $x$-axis, we use the disk method:

$V = \pi \int_{0}^{\pi} (\cos 4x)^2 \, dx$

Using the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$, we have:

$V = \pi \int_{0}^{\pi} \frac{1 + \cos(8x)}{2} \, dx$

Calculating the integral:

$V = \frac{\pi}{2} \left[ x + \frac{\sin(8x)}{8} \right]_{0}^{\pi} = \frac{\pi}{2} \left[ \pi + 0 - (0) \right] = \frac{\pi^2}{2}$

Thus, the volume of the solid formed by the rotation is:

$$\frac{\pi^2}{2} \text{ cubic units}.$

Starting with the acceleration given by:

$\ddot{x} = 2 - \frac{x}{2}$

Acceleration can also be expressed using velocity:

$\ddot{x} = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$

Setting the two expressions for acceleration equal, we have:

$v \frac{dv}{dx} = 2 - \frac{x}{2}$

Separating variables yields:

$vdv = \left(2 - \frac{x}{2}\right)dx$

Integrating both sides leads to:

$\frac{1}{2} v^2 = 2x - \frac{x^2}{4} + C$

Using the boundary condition $v = 4$ when $x = 0$, we find:

ightarrow C = 8$$ Thus, we can express: $$\frac{1}{2} v^2 = 2x - \frac{x^2}{4} + 8$$ Therefore: $$v^2 = 4x - \frac{x^2}{2} + 16$$