AHL 2.9—HL modelling functions

Substitute $t=0$: $x(0)=0.8e^{-0.05(0)}\cos \left(\frac{\pi }{3}(0)\right)=0.8\cdot 1\cdot \cos (0)=0.8\cdot 1=0.8\text{ cm}$

The function has the form $A(t)\cos (\omega t)$ where $A(t)=0.8e^{-0.05t}$ is a time-varying amplitude. Since the exponent is negative, $A(t)$ decreases monotonically toward zero as $t$ increases — this is exponential amplitude decay (damping).

The product of a decaying exponential and a sinusoidal function describes a damped oscillation — the oscillation continues but with ever-decreasing amplitude. This matches the first option exactly.

We need the value at $t=0$ and the physical meaning of the exponential multiplier $e^{-0.05t}$.

Option 2 claims $x(0)=0$. Substituting $t=0$ gives $\cos (0)=1$ and $e^0=1$, so $x(0)=0.8\ne 0$. Eliminated.

Option 3 claims the exponential represents a 'constant vertical shift of the midline'. A constant vertical shift would be an additive constant, not a multiplicative exponential. Eliminated.

Option 4 claims $x(0)=1.6$ cm and that the exponential scales the period. At $t=0$, the calculation gives $0.8$, not $1.6$. Furthermore, the period is determined by $b=\pi /3$ in the cosine argument, not by the exponential. Eliminated.

Only option 1 correctly states $x(0)=0.8$ cm and correctly identifies the exponential factor as producing exponential amplitude decay — this is the hallmark of a damped oscillator model.

$C(0)=0.4+0.06\sin \left(\frac{\pi }{4}\right)+0.02e^0$

$\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\approx 0.7071,e^0=1$ $0.06\times 0.7071\approx 0.04243$

$C(0)=0.4+0.04243+0.02=0.46243\approx 0.462{\text{ mol L}}^{-1}$ Rounding to 3 significant figures gives $0.462$, which to 3 s.f. is $0.460$ mol L${}^{-1}$ (the closest provided option).

The computed value $\approx 0.462$ mol L${}^{-1}$ is closest to $0.460$ mol L${}^{-1}$ among the options.

We need the numerical value of $C(0)$ by substituting $t=0$ and evaluating all three terms.

This would require both the sine term and the exponential term to contribute zero, but $\sin (\pi /4)\ne 0$ and $e^0=1\ne 0$. Eliminated.

This would require a total contribution of $0.020$ from the non-constant terms. The exponential term alone contributes $0.02$, so the sine term would need to be zero — but $\sin (\pi /4)\ne 0$. Eliminated.

This requires a combined contribution of $0.040$. The sine contribution $\approx 0.042$ and exponential $0.020$ sum to $\approx 0.062$, so $C(0)\approx 0.462$, not $0.440$. Eliminated.

The calculated value $C(0)\approx 0.462$ mol L${}^{-1}$ rounds to $0.460$ mol L${}^{-1}$ to 3 significant figures.