Réitigh T=2pisqrt{l/98}quad[2]

$$T = 2 \pi \sqrt { \frac { l } { 98 } } \quad [ 2 ]$$

Réitigh do l.

$$l=\frac{49\times \left(\frac{T}{\pi }\right)^{2}}{8}$$

$$T\geq 0$$

Céimeanna chun cothromóidí fréimhe a réiteach

Méadaigh 2 agus 2 chun 4 a fháil.

$$T=4\pi \sqrt{\frac{l}{98}}$$

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$4\pi \sqrt{\frac{l}{98}}=T$$

Roinn an dá thaobh faoi 4\pi .

$$\frac{4\pi \sqrt{\frac{1}{98}l}}{4\pi }=\frac{T}{4\pi }$$

Má roinntear é faoi 4\pi cuirtear an iolrúchán faoi 4\pi ar ceal.

$$\sqrt{\frac{1}{98}l}=\frac{T}{4\pi }$$

Cearnaigh an dá thaobh den chothromóid.

$$\frac{1}{98}l=\frac{T^{2}}{16\pi ^{2}}$$

Iolraigh an dá thaobh faoi 98.

$$\frac{\frac{1}{98}l}{\frac{1}{98}}=\frac{T^{2}}{\frac{1}{98}\times 16\pi ^{2}}$$

Má roinntear é faoi \frac{1}{98} cuirtear an iolrúchán faoi \frac{1}{98} ar ceal.

$$l=\frac{T^{2}}{\frac{1}{98}\times 16\pi ^{2}}$$

Roinn \frac{T^{2}}{16\pi ^{2}} faoi \frac{1}{98} trí \frac{T^{2}}{16\pi ^{2}} a mhéadú faoi dheilín \frac{1}{98}.

$$l=\frac{49T^{2}}{8\pi ^{2}}$$

Réitigh do T.

$$T=\frac{2\pi \sqrt{2l}}{7}$$

$$l\geq 0$$