As you know, when calculating the period of oscillation of a mathematical pendulum, you substitute $\sin \alpha$ for $\alpha$, where $\alpha$ is the angle of deviation of the pendulum string from the vertical. Find out whether the period calculated in this way is smaller or larger than the real period. A mathematical pendulum is a point mass $m$ suspended on a weightless, inextensible thread of length $l$.
Decision:
When the thread is deflected from the vertical by an angle $\alpha$, the point mass $m$ follows a path of length $s = l \alpha$ along the arc of a circle with radius $l$. For the projection of acceleration on the direction of the tangent to the trajectory at time $t$, when the thread makes an angle $\alpha$ with the vertical, we have
$\alpha = \frac{d^{2}s}{dt^{2}}=l \frac{d^{2} \alpha}{dt^{2}}$.
Only the force of gravity $mg$ has a component on the tangent direction. At time $t$, this component is equal to
$F = - mg \sin \alpha$.
According to Newton's law II we obtain
$m_{l} \frac{d^{2} \alpha}{dt^{2}} = -mg \sin \alpha$. (1)
When considering small oscillations ($\alpha \ll 1$), equation (1) usually assumes $\sin \alpha \approx \alpha$ in equation (1), and it takes the form of
$\frac{d^{2} \alpha}{dt^{2}} = - \frac{g \alpha}{l}$ (2)
(harmonic oscillation equation). Since $| \sin \alpha | < | \alpha | $, it is clear, that by replacing $\sin \alpha$ with $\alpha$; we increase the right-hand side, and hence the left-hand side of equation (1). By increasing the angular acceleration at each point of the trajectory $d^{2} \alpha / dt^{2}$, we also increase the angular velocity $d \alpha / dt$. Therefore, the actual period of oscillation with a given amplitude is larger than the value $T = 2 \pi \sqrt{l/g}$ obtained from the approximate equation (2).
Decision:
When the thread is deflected from the vertical by an angle $\alpha$, the point mass $m$ follows a path of length $s = l \alpha$ along the arc of a circle with radius $l$. For the projection of acceleration on the direction of the tangent to the trajectory at time $t$, when the thread makes an angle $\alpha$ with the vertical, we have
$\alpha = \frac{d^{2}s}{dt^{2}}=l \frac{d^{2} \alpha}{dt^{2}}$.
Only the force of gravity $mg$ has a component on the tangent direction. At time $t$, this component is equal to
$F = - mg \sin \alpha$.
According to Newton's law II we obtain
$m_{l} \frac{d^{2} \alpha}{dt^{2}} = -mg \sin \alpha$. (1)
When considering small oscillations ($\alpha \ll 1$), equation (1) usually assumes $\sin \alpha \approx \alpha$ in equation (1), and it takes the form of
$\frac{d^{2} \alpha}{dt^{2}} = - \frac{g \alpha}{l}$ (2)
(harmonic oscillation equation). Since $| \sin \alpha | < | \alpha | $, it is clear, that by replacing $\sin \alpha$ with $\alpha$; we increase the right-hand side, and hence the left-hand side of equation (1). By increasing the angular acceleration at each point of the trajectory $d^{2} \alpha / dt^{2}$, we also increase the angular velocity $d \alpha / dt$. Therefore, the actual period of oscillation with a given amplitude is larger than the value $T = 2 \pi \sqrt{l/g}$ obtained from the approximate equation (2).