A particle moving along $y$-axis has an acceleration $Fy$ towards the origin, where $F$ is a positive and even function of $y$. The periodic time when the particle vibrates between $y=-a$ and $y=a$ is $T$. Show that $$\frac{2\pi}{\sqrt{F_1}}<T<\frac{2\pi}{\sqrt{F_2}}$$ where $F_1$ and $F_2$ are the greatest and the least values of $F$ within the range $[-a,a]$. Further show that when a simple pendulum of length $l$ oscilates through $30°$ on either side of the verical line, $T$ lies beween $2\pi\sqrt{\frac lg}$ and $2\pi\sqrt{\frac lg}\sqrt{\frac{\pi}3}$.
So, for the first part, if $y(t)$ is path under this motion, then we have $\dfrac{d^2y}{dt^2}= -F(y)y$ where $F(y) > 0$ for all $y$ and $F(y)=F(-y)$.
Comparing with simple harmonic motion, the claim of the question seemed obvious, that $T$ should indeed lie between $\frac{2\pi}{\sqrt{F_1}}$ and $\frac{2\pi}{\sqrt{F_2}}$, but then I realised that under SHM, things are quite easy as there time required to go to center of oscillation from the farthest point attainable under oscillation is same as the reverse journey, i.e., from center to the farthest point. But, under this general problem, this does not seem to be the case - here acceleration $F_1$ should slow the particle down while acceleration $F_2$ should speed it up in the reverse journey from center to farthest point.
In the second part involving simple pendulum, I can't treat this as a simple harmonic motion, as I don't have luxury of approximating $\sin \theta ≈ \theta$ (as angle involved here is not very small - like less than $15°$). I think that problem wants me to use first part here and find out suitble $F$ so that the bounds of time period $T$ can be calculated.
So, how can I proceed further?
As is well known, the first integral of the equation of motion $$ \ddot{y} = -F(y)y \tag{1} $$ is the equation of conservation of mechanical energy, $$ \frac{1}{2}\dot{y}^2 + U(y) = E = U(a), \tag{2}$$ where $$U(y)=\int_0^yF(x)x\,dx \tag{3}$$ is the potential energy associated with the force $-F(y)y$.
Since the time spent to go from $-a$ to $a$ is the same as the time to go from $a$ to $-a$, we can use $(2)$ to calculate the period $T$ as $$ T=2\int_{-a}^{a}\frac{dy}{\dot{y}}=2\int_{-a}^{a}\frac{dy}{\sqrt{2(U(a)-U(y))}} =4\int_{0}^{a}\frac{dy}{\sqrt{2(U(a)-U(y))}}, \tag{4} $$ where in the last equality we used the fact that $U(-y)=U(y)$ (this follows from $(3)$ and the fact that $F(-y)=F(y)$). Now we use the inequalities $F_2\leq F(y)\leq F_1$ in the interval $[0,a]$ to derive a lower and an upper bound for $T$: \begin{align} F_2\leq F\leq F_1 &\Rightarrow \int_y^a F_2x\,dx \leq \int_y^a F(x)x\,dx \leq \int_y^a F_1x\,dx \\ &\Rightarrow \frac{1}{2}F_2(a^2-y^2) \leq U(a)-U(y) \leq \frac{1}{2}F_1(a^2-y^2) \\ &\Rightarrow \int_0^{a}\frac{dy}{\sqrt{F_2(a^2-y^2)}} \geq \int_{0}^{a}\frac{dy}{\sqrt{2(U(a)-U(y))}} \geq \int_0^{a}\frac{dy}{\sqrt{F_1(a^2-y^2)}} \\ &\Rightarrow \frac{\pi}{2\sqrt{F_2}} \geq \frac{T}{4} \geq \frac{\pi}{2\sqrt{F_1}} \\ &\Rightarrow \frac{2\pi}{\sqrt{F_1}} \leq T \leq \frac{2\pi}{\sqrt{F_2}}. \tag{5} \end{align}
In order to apply the inequalities $(5)$ to the period of the simple pendulum, we note that its equation of motion is given by$^{(*)}$ $$ \ddot{\theta}=-\frac{g}{l}\sin\theta = -F(\theta)\theta, \tag{6} $$ where $$ F(\theta)=\frac{g}{l}\frac{\sin \theta}{\theta}. \tag{7} $$ The maximum and the minimum values attained by $F(\theta)$ in the interval $[-30^{\circ},30^{\circ}]$ (or $[-\frac{\pi}{6},\frac{\pi}{6}]$, when angles are expressed in radians) are, respectively, $$ F_1=F(0)=\frac{g}{l}\quad{\rm and}\quad F_2=F\left(\frac{\pi}{6}\right)=\frac{3g}{\pi l}. \tag{8} $$ Inserting these values in $(5)$ shows that the period $T$ of the simple pendulum when its angular amplitude is $30^{\circ}$ lies between $2\pi\sqrt{\frac{l}{g}}$ and $2\pi\sqrt{\frac{\pi}{3}\frac{l}{g}}$.
$^{(*)}$ https://en.wikipedia.org/wiki/Pendulum_(mechanics)