Calculating the time for current reducing to zero from a charged inductor in a series LC circuit
An idealized model during the discharge, when the output diode is on, is a simple LC circuit. The solution to that model for the voltage and current are the sine and cosine resulted from the 2nd order differential equation. so you can use the sine and cosine solution fitted with initial conditions for the calculation.

But in practice, for a boost circuit, you don't see people using the LC model for this calculation. The reason is, the output capacitor is specifically chosen so that the output voltage is near constant (usually less than 1% ripple).

It becomes much simpler with the output voltage (Vout) assumed to be constant:
$$ L \frac{dI_L}{dt} = V_L = V_{in} - V_{out} = constant $$
Integrate and fit chosen initial conditions, you get a simple negative ramp:
$$ I_L(t) = I_{Lpeak} - \frac{1}{L} (V_{out} - V_{in}) t $$
The typical way of using this relationship for boost circuit design is:
$$ I_{Lpeak} = \frac{1}{L} (V_{out} - V_{in}) t_{discharge} $$
\$t_{discharge}\$ is the time it takes for the current \$I_{L}\$ to discharge from \$I_{Lpeak}\$ to 0.

When \$I_L\$ reaches 0, the output diode turns off. \$I_L\$ stays 0 and \$V_L\$ also drops to 0. (Therefore, there is no oscillation.)