Soal Latihan 10.1 dan 10.2 (Gelombang Bunyi - Sagufindo Kinarya)

Soal Latihan 10.1
Berapakah frekuensi nada dasar dan nada atas pertama suatu pipa organa terbuka yang panjangnya 1 meter bila cepat rambat bunyi di udara 344 m/s?
Berapakah frekuensi-frekuensi tersebut bila pipa organa itu tertutup?

Diketahui :

$$\begin{aligned} n &= 0 \nonumber\\ n &= 1 \nonumber\\ L_{POB} &= 1 \ m \nonumber\\ L_{POT} &= 1 \ m \nonumber\\ v &= 344 \ m/s \end{aligned}$$

Ditanya :
$$\begin{aligned} f_{0 \ POB} &= …? \nonumber\\ f_{1 \ POB} &= …? \nonumber\\ f_{0 \ POT} &= …? \nonumber\\ f_{1 \ POT} &= …? \nonumber\\ \end{aligned}$$

Jawab:

$$\begin{aligned} f_{0 \ POB} &= \frac{(n+1) \times v}{2 \times L_{POB}} \nonumber\\ &= \frac{(0+1)\times 344}{2 \times 1 } \nonumber\\ &= \frac{344}{2} \nonumber\\ &= 172 \ Hz \end{aligned}$$ $$\begin{aligned} f_{1 \ POB} &= \frac{(n+1) \times v}{2 \times L_{POB}} \nonumber\\ &= \frac{(1+1)\times 344}{2 \times 1 } \nonumber\\ &= \frac{2 \times 344}{2} \nonumber\\ &= 344 \ Hz \end{aligned}$$ $$\begin{aligned} f_{0 \ POT} &= \frac{(2n+1) \times v}{4 \times L_{POB}} \nonumber\\ &= \frac{(2.0+1)\times 344}{4 \times 1 } \nonumber\\ &= \frac{344}{4} \nonumber\\ &= 86 \ Hz \end{aligned}$$ $$\begin{aligned} f_{1 \ POT} &= \frac{(2n+1) \times v}{4 \times L_{POB}} \nonumber\\ &= \frac{(2.1+1)\times 344}{4 \times 1 } \nonumber\\ &= \frac{3 \times 344}{4} \nonumber\\ &= \frac{1032}{4} \nonumber\\ &= 258 \ Hz \end{aligned}$$  

Baca Juga : Pembahasan Uji Prestasi Mandiri 10.1 Gelombang Bunyi - Sagufindo Kinarya

Soal Latihan 10.2
Sebuah pipa organa terbuka menghasilkan nada atas ketiga dengan frekuensi 1700 Hz. Cepat rambat bunyi di udara 340 m/s.
a.    Berapa panjang pipa organa tersebut?
b.    Jika pipa organa tersebut tertutup, berapa panjang pipa yang digunakan?

Diketahui :

$$\begin{aligned} n &=3 \nonumber\\ f_{3 \ POB} &= 1700 \ Hz \nonumber\\ v &= 340 \ m/s \end{aligned}$$

Ditanya :
$$\begin{aligned} a.    L_{POB} \nonumber\\ b.    L_{POT} \end{aligned}$$

Jawab:

$$\begin{aligned} f_{3 \ POB} &= \frac{(n+1) \times v}{2 \times L_{POB}} \\ 1700 &= \frac{(3+1) \times 340}{2 \times L_{POB}} \nonumber\\ L_{POB} &= \frac{4 \times 340}{2 \times 1700} \nonumber\\ &= 0,4 \ meter \nonumber\\ &= 40 \ cm \end{aligned}$$ $$\begin{aligned}f_{3 \ POT} &= \frac{(2n+1) \times v}{4 \times L_{POT}}\\ 1700 &= \frac{(2.3+1) \times 340}{4 \times L_{POT}}\\ L_{POB} &= \frac{7 \times 340}{4 \times 1700}\\ &= 0,35 \ meter\\ &= 35 \ cm \end{aligned}$$