Tugas Fisika Matematika 2

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Tugas Fisika Matematika 2 Nama : Fidya Alfitri Nim

: H21115504

Mod 2 2. A metal plate covering the first quadrant has the edge which is along the y axis insulated and the edge which is along the x axis held at 𝑒(π‘₯, 0) = {

100(2 βˆ’ π‘₯), π‘“π‘œπ‘Ÿ 0 < π‘₯ < 2, 0 , π‘“π‘œπ‘Ÿ π‘₯ < 2

Find the steady-state temperature distribution as a function of x and y. Hint: Follow the procedure of Example 2, but use a cosine transform (because βˆ‚u/βˆ‚x = 0 for x = 0). Leave your answer as an integral like (9.13).

Jawab: 𝑒(π‘₯, 0) = {

100(2 βˆ’ π‘₯), π‘“π‘œπ‘Ÿ 0 < π‘₯ < 2, 0 , π‘“π‘œπ‘Ÿ π‘₯ < 2

Distribusi suhu steady state sebagai fungsi dari x dan y. Maka solusinya akan menjadi: 𝑒(π‘₯, 0) = {

𝑒 π‘˜π‘¦ , π‘“π‘œπ‘Ÿ 0 < π‘₯ < 2, 𝑒 βˆ’π‘˜π‘¦ , π‘“π‘œπ‘Ÿ π‘₯ < 2

Karena merupakan kuadran pertama, maka arah y, dengan u β†’ 0 dan y β†’ ∞. 𝑒 π‘˜π‘¦ dapat diabaikan, dengan

πœ•π‘’ πœ•π‘₯

= 0 dan x = 0. Sumbu y dengan sin (kx) diabaikan.

sehingga solusinya berbentuk: ∞

𝑒 = ∫ 𝐡(π‘˜)𝑒 βˆ’π‘˜π‘₯ cos(π‘˜π‘₯)π‘‘π‘˜ 0

Untuk y = 0 maka dapat dituliskan: ∞

𝑒(π‘₯, 0) = ∫ 𝐡(π‘˜) cos(π‘˜π‘₯)π‘‘π‘˜ 0 ∞

2 𝑓𝑐 (π‘₯) = √ ∫ 𝑔𝑐 (π‘˜) cos(π‘˜π‘₯)π‘‘π‘˜ πœ‹ 0

Dengan mencocokkan dengan transformasi Fourier kosinus maka dapat dituliskan: 2 𝐡(π‘˜) = √ 𝑔𝑐 (π‘˜) πœ‹ Menerapkan definisi sepasang transformasi Fourier kosinus menggunakan konvensi, B (k) dapat dihitung sebagai berikut: ∞

2 𝐡(π‘˜) = √ ∫ 𝑒(π‘₯, 0) cos(π‘˜π‘₯)π‘‘π‘˜ πœ‹ 0

∞

200 =√ ∫ (2 βˆ’ π‘₯) cos(π‘˜π‘₯)π‘‘π‘˜ πœ‹ 0

400 𝑠𝑖𝑛2 (π‘˜) √ = πœ‹ π‘˜2 Sehingga didapatkan solusinya yaitu: ∞

400 𝑠𝑖𝑛2 (π‘˜) 𝑒(π‘₯, 0) = √ ∫ cos(π‘˜π‘₯)π‘‘π‘˜ πœ‹ π‘˜2 0

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