Q.Deduce the expression for period of simple pendulum. Hence state the factors on which its period depends.

[!TLDR] Small-angle restoring force gives acceleration-per-unit-displacement g/L, so $T=2\pi\sqrt{L/g}$; depends on L and g only (not mass or amplitude, fo …

See Answer-in-Brief (iii)/(iv) derivations for the full steps. In summary: for a bob displaced through small angle $\theta$, the restoring force $F=-mg\sin\theta\approx-\frac{mgx}{L}$ (using $\sin\theta\approx\theta\approx x/L$), giving acceleration $a=-\frac{g}{L}x$ -- proportional to and opposite to displacement x, i.e. S.H.M., with acceleration-per-unit-displacement $g/L$. Substituting into $T=2\pi\sqrt{\text{displacement}/\text{acceleration per unit displacement}}$ gives $T=2\pi\sqrt{L/g}$ (Eq. 5.28). The factors the period depends on, read directly off this formula: the string length L (period $\propto\sqrt L$) and the local acceleration due to gravity g (period $\propto1/\sqrt g$ …

Derive the small-angle restoring force, extract g/L as the acceleration-per-unit-displacement, substitute into the period template, then re …

Listing amplitude or mass as factors the period depends on, when in fact (for small os …