The equation of motion for a particle moving in simple harmonic motion is given by d^2x/dt^2 = -n^2 x, where n is a positive constant, x is the displacement of the particle and t is time - HSC - SSCE Mathematics Extension 1 - Question 5 - 2009 - Paper 1

The maximum speed of the particle occurs when the displacement is at its maximum. Since the speed is highest when the particle passes through the equilibrium position (where x = 0), we can substitute 'x' into our earlier equation:

$v_{max} = n a$

Thus, the maximum speed is given by the product of the constant 'n' and the amplitude 'a'.

The maximum acceleration can be found by evaluating the acceleration derived from the equation:

$\frac{d^2x}{dt^2} = -n^2 x$

The maximum acceleration occurs at maximum displacement, which is:

$a_{max} = n^2 a$

This shows that the maximum acceleration is proportional to the amplitude and the square of the constant 'n'.

For simple harmonic motion, the position can be expressed as:

$x(t) = a imes ext{sin}(nt)$

The speed can be derived to be:

$v(t) = \frac{dx}{dt} = an \times ext{cos}(nt)$

Setting the speed equal to half the maximum speed:

$\frac{an}{2} = an \times ext{cos}(nt)$

which simplifies to:

$\text{cos}(nt) = \frac{1}{2}$

The first time this occurs is when:

$nt = \frac{\pi}{3} ext{ or } \frac{5\pi}{3},$ leading to the first instance:

$t = \frac{\pi}{3n}$.

The volume of the water in the tank can be determined from the cross-sectional area at depth 'h'. Since the cross-section is an isosceles triangle, the formula for the area A can be integrated.

Therefore, we have:

$V_t = A \times L,$ with 'A' being the area of the triangle and 'L' being the length of the tank, leading to:

$V_t = \frac{1}{2} \times base \times height \times length.$

This can be evaluated correctly with dimensions based on the height 'h'.

To find the area of the top surface when the depth of water is 'h', we can evaluate the dimensions relative to the triangle. From the geometry of the isosceles triangle, we find:

$A = \frac{1}{2} \times base(h) \times height(h)$

This can be rearranged into the form required:

$A = 20 \times \frac{3}{h}$.

From the rate of evaporation formula, given by:

$\frac{dV}{dt} = -kA$

where 'k' is a constant determined experimentally. By substituting A from part (ii), we find:

$\frac{dV}{dt} = -k \times \left(20 \times \frac{3}{h}\right)$

Using the relationship of volume to height, we can express:

$\frac{dV}{dt} = A \frac{dh}{dt}$, leading to:

$\frac{dh}{dt} = -\frac{kA}{Volume(h)}$.

Using the rate of evaporation determined from earlier parts, take the times for the height change from 3 m to 2 m (100 days) as a baseline. By using proportional relationships, if evaporation is linear, apply the found proportionality constant to determine the time taken to drop from 2 m to 1 m by integrating the rate of change over that interval.