$3.1.16.$ Determine within what limits the tension force of a mathematical pendulum is if the amplitude of oscillations $x_0$ is much smaller than the length of the string $l$, the mass of the pendulum is $m$.
We will find the tension force of the thread as $$ T_\text{min}=mg\cos\varphi $$ Let's use the approximation, for $\varphi\ll1$, $\cos\varphi=1-\frac{\varphi^2}{2}\quad(1)$: $$ mg\cos\varphi \approx mg\left(1-\frac{\varphi^2}{2}\right) $$ $$ \boxed{T_\text{min} =mg\left(1-\frac{x_0^2}{2l^2}\right)} $$ Let's write down Newton's second law $$ ma=T_\text{max}-mg $$ $$ T_\text{max}=m\left(\frac{\upsilon^2}{l}+g\right) $$ Also, let's write down the law of conservation of energy $$ mgl(1-\cos\varphi)=\frac{m\upsilon^2}{2} $$ $$ \upsilon^2=2gl(1-\cos\varphi) $$ Given the approximation $(1)$ $$ v^2\approx gl\varphi^2=\frac{gx_0^2}{l} $$ Let's substitute this expression for speed into the expression for $T_\text{max}$: $$ \boxed{T_\text{max}=mg\left(1+\frac{x_0^2}{l^2}\right)} $$
$$mg\left(1-\frac{x_0^2}{2l^2}\right) < T < mg\left(1+\frac{x_0^2}{l^2}\right)$$