Whakaoti i te T=2pisqrt{l/98}quad[2]

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

Whakaoti mō l

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

$$T\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

Whakareatia te 2 ki te 2, ka 4.

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

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

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

Whakawehea ngā taha e rua ki te 4\pi .

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

Mā te whakawehe ki te 4\pi ka wetekia te whakareanga ki te 4\pi .

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

Pūruatia ngā taha e rua o te whārite.

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

Me whakarea ngā taha e rua ki te 98.

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

Mā te whakawehe ki te \frac{1}{98} ka wetekia te whakareanga ki te \frac{1}{98}.

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

Whakawehe \frac{T^{2}}{16\pi ^{2}} ki te \frac{1}{98} mā te whakarea \frac{T^{2}}{16\pi ^{2}} ki te tau huripoki o \frac{1}{98}.

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

Whakaoti mō T (complex solution)

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

Whakaoti mō l (complex solution)

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

$$|\frac{arg(T^{2})}{2}-arg(T)|<\pi \text{ or }T=0$$

Whakaoti mō T

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

$$l\geq 0$$