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.

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    <h3 id="back-link"><a href="/#3.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $3.1.16.$ 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 $m$.
    </p>

    <h3>Solution</h3>
    <p>
        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 $T_\text{max}$:
$$\boxed{T_\text{max}=mg\left(1+\frac{x_0^2}{l^2}\right)}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$mg\left(1-\frac{x_0^2}{2l^2}\right) < T < mg\left(1+\frac{x_0^2}{l^2}\right)$$
    </p>
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