Whakaoti i te T=2pisqrt{L/a}

$$T = 2 \pi \sqrt { \frac { L } { a } }$$

Whakaoti mō L

$$L=\frac{a\times \left(\frac{T}{\pi }\right)^{2}}{4}$$

$$T\geq 0\text{ and }a\neq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$2\pi \sqrt{\frac{L}{a}}=T$$

Whakawehea ngā taha e rua ki te 2\pi .

$$\frac{2\pi \sqrt{\frac{1}{a}L}}{2\pi }=\frac{T}{2\pi }$$

Mā te whakawehe ki te 2\pi ka wetekia te whakareanga ki te 2\pi .

$$\sqrt{\frac{1}{a}L}=\frac{T}{2\pi }$$

Pūruatia ngā taha e rua o te whārite.

$$\frac{1}{a}L=\frac{T^{2}}{4\pi ^{2}}$$

Whakawehea ngā taha e rua ki te a^{-1}.

$$\frac{\frac{1}{a}La}{1}=\frac{T^{2}}{4\pi ^{2}\times \frac{1}{a}}$$

Mā te whakawehe ki te a^{-1} ka wetekia te whakareanga ki te a^{-1}.

$$L=\frac{T^{2}}{4\pi ^{2}\times \frac{1}{a}}$$

Whakawehe \frac{T^{2}}{4\pi ^{2}} ki te a^{-1}.

$$L=\frac{aT^{2}}{4\pi ^{2}}$$

Whakaoti mō T

$$T=2\pi \sqrt{\frac{L}{a}}$$

$$\left(L\geq 0\text{ and }a>0\right)\text{ or }\left(L\leq 0\text{ and }a<0\right)$$