Frekuensi = f = 0,2 detik

Frekuensi sudut = ω = 2πf = 0,4π

Bilangan gelombang = k = \(\dfrac{2 \pi}{\lambda} =  \dfrac{2 \pi}{10} = 0,2 \pi\)

Persamaan gelombang:

\begin{equation*}
y = 2 \sin (0,4 \pi t - 0,2 \pi x)
\end{equation*}

A. Persamaan gelombang

Panjang gelombang

\begin{equation*}
\begin{split}
v & = \lambda \:.\: f \\\\
4 & = \lambda \:.\: 0,5 \\\\
\lambda & = 8 \text{ m}
\end{split}
\end{equation*}

\begin{equation*}
\begin{split}
y & = A \sin (\omega t - kx + \theta_o) \\\\
y & = A \sin (2 \pi f t - \frac{2 \pi}{\lambda} x + \theta_o) \\\\
y & = 10 \sin (2 \pi \:.\: 0,5 \:.\: t - \frac{2 \pi}{8} x + \theta_o) \\\\
y & = 10 \sin (\pi t - \frac{\pi}{4} x + \theta_o) \text{ cm}
\end{split}
\end{equation*}

Menentukan \(\theta_o\)

Simpangan mula-mula y_o = 5 cm (t_o = 0, x_o = 0)

\begin{equation*}
\begin{split}
y & = 10 \sin (\pi t - \frac{\pi}{4} x + \theta_o) \\\\
5 & = 10 \sin \theta_o \\\\
\frac{1}{2} & = \sin \theta_o \\\\
\theta_o & = 30^{\text{o}} \\\\
\theta_o & = \frac{\pi}{6}
\end{split}
\end{equation*}

Persamaan gelombang

\begin{equation*}
y = 10 \sin (\pi t - \frac{\pi}{4} x + \frac{\pi}{6}) \text{ cm}
\end{equation*}

B. Simpangan, kecepatan dan percepatan pada x = 4 m dan t = 1 detik

Simpangan

\begin{equation*}
\begin{split}
y & = 10 \sin (\pi t - \frac{\pi}{4} x + \frac{\pi}{6}) \\\\
y & = 10 \sin (\pi (1) - \frac{\pi}{4} (4) + \frac{\pi}{6}) \\\\
y & = 10 \sin \frac{\pi}{6} \\\\
y & = 10 \:.\: \frac{1}{2} \\\\
y & = 5 \text{ cm}
\end{split}
\end{equation*}

Kecepatan getar

\begin{equation*}
\begin{split}
v_y & = \frac{dy}{dt} = \omega A \cos (\omega t - kx + \theta_o) \\\\
v_y & = 10 \pi \cos (\pi t - \frac{\pi}{4} x + \frac{\pi}{6}) \\\\
v_y & = 10 \pi \cos (\pi (1) - \frac{\pi}{4} (4) + \frac{\pi}{6}) \\\\
v_y & = 10 \pi \:.\: \cos \frac{\pi}{6} \\\\
v_y & = 10 \pi \:.\: \frac{1}{2} \sqrt{3} \\\\
v_y & = 5 \sqrt{3} \text{ cm/s}
\end{split}
\end{equation*}

Percepatan getar

\begin{equation*}
\begin{split}
a_y & = - \omega^2 y \\\\
a_y & = \pi^2 \:.\: 5 \\\\
a_y & = 5 \pi^2 \text{ cm/s}^2
\end{split}
\end{equation*}

C. Sudut fase pada x = 2 m dan t = 0,5 detik

\begin{equation*}
\begin{split}
\theta & = \pi t - \frac{\pi}{4} x + \frac{\pi}{6} \\\\
\theta & = \pi (0,5) - \frac{\pi}{4} (2) + \frac{\pi}{6} \\\\
\theta & = \frac{\pi}{6} \\
\end{split}
\end{equation*}

D. Beda fase dari dua lokasi yang berjarak 20 m

\begin{equation*}
\begin{split}
\Delta \varphi & = \frac{\Delta x}{\lambda} \\\\
\Delta \varphi & = \frac{20}{8} \\\\
\Delta \varphi & = 2\frac{1}{2} \\\\
\Delta \varphi & = \frac{1}{2}
\end{split}
\end{equation*}

E. Beda fase suatu lokasi dengan selisih waktu 4 detik

\begin{equation*}
\begin{split}
\Delta \varphi & = \frac{\Delta t}{T} \\\\
\Delta \varphi & = \frac{4}{2} \\\\
\Delta \varphi & = 2
\end{split}
\end{equation*}

A. Frekuensi, amplitudo dan cepat rambat gelombang

\begin{equation*}
\begin{split}
y & = 2 \sin \pi \left(\frac{1}{3}x - \frac{1}{2}t + \frac{1}{4}\right) \\\\
y & = 2 \sin \left(\frac{1}{3} \pi x - \frac{1}{2} \pi t + \frac{1}{4} \pi \right) \\\\
A & = 2 \text{ cm} \\\\
\omega & = \frac{1}{2} \pi \\\\
2 \pi f & = \frac{1}{2} \pi \\\\
f & =  \frac{1}{4} \text{ Hz} \\\\
k & = \frac{2 \pi}{\lambda} \\\\
\frac{1}{3} \pi & = \frac{2 \pi}{\lambda} \\\\
\lambda & = 6 \text{ m} \\\\
v & = \lambda \:.\: f \\\\
v & = 6 \:.\: \frac{1}{4} \\\\
v & = 1,5 \text{ m/s}
\end{split}
\end{equation*}

B. Beda fase antara dua tempat yang berjarak 48 m

\begin{equation*}
\begin{split}
\Delta \varphi & = \frac{\Delta x}{\lambda} \\\\
\Delta \varphi & = \frac{48}{6} \\\\
\Delta \varphi & = 8 \\\\
\Delta \varphi & = 0
\end{split}
\end{equation*}

Kedua tempat sefase

Panjang gelombang

\begin{equation*}
\begin{split}
v & = \lambda \:.\: f \\\\
60 & = \lambda \:.\: 100 \\\\
\lambda & = 0,6 \text{ m}
\end{split}
\end{equation*}

A. Jarak dua titik jika beda fasenya 1/3

\begin{equation*}
\begin{split}
\Delta \varphi & = \frac{\Delta x}{\lambda} \\\\
\frac{1}{3} & = \frac{\Delta x}{0,6} \\\\
\Delta x & = 0,2 \text{ m}
\end{split}
\end{equation*}

B. Beda fase pada satu titik jika selisih waktu 0,012 detik

\begin{equation*}
\begin{split}
\Delta \varphi & = \frac{\Delta t}{T} \\\\
\Delta \varphi & = \frac{0,012}{0,01} \\\\
\Delta \varphi & = 1,2 \\\\
\Delta \varphi & = 0,2
\end{split}
\end{equation*}

A. Persamaan gelombang stationer ujung tetap

\begin{equation*}
\begin{split}
y & = 2A \sin kx  \cos \omega t \\\\
y & = 2A \sin \tfrac{2\pi}{\lambda} x  \cos 2 \pi f t \\\\
y & = 2 \:.\: 50 \sin \tfrac{2\pi}{2} x  \cos 2 \pi \:.\: 2 t \\\\
y & = 100 \sin \pi x  \cos 4 \pi t \text{ cm}
\end{split}
\end{equation*}

B. Amplitudo gelombang stasioner pada x = 2 meter 

\begin{equation*}
\begin{split}
A_s & = 100 \sin \pi x \\\\
A_s & = 100 \sin \pi (2) \\\\
A_s & = 0
\end{split}
\end{equation*}

C. Simpangan pada  x = 0,25 meter dan t = 4 detik

\begin{equation*}
\begin{split}
y & = 100 \sin \pi x  \cos 4 \pi t \text{ cm} \\\\
y & = 100 \sin \pi (0,25)  \cos 4 \pi (4) \\\\
y & = 100 \sin 0,25 \pi \cos 16 \pi \\\\
y & = 100 \:.\: \tfrac{1}{2} \sqrt{2} \:.\: \cos 0 \\\\
y & = 100 \:.\: \tfrac{1}{2} \sqrt{2} \:.\: (1) \\\\
y & = 50 \sqrt{2}
\end{split}
\end{equation*}

D. Jarak antara dua simpul yang berdekatan

\begin{equation*}
\begin{split}
x & = (n - 1) \: \frac{\lambda}{2} \\\\
x & = (n - 1) \: \frac{2}{2} \\\\
x & = (n - 1) \text{ meter} \\\\
n = 1 & \rightarrow x = 0 \\\\
n = 2 & \rightarrow x = 1 \text{ meter}
\end{split}
\end{equation*}

Jarak dua simpul berdekatan adalah 1 meter.

E. Jarak perut kedua dari ujung tetap

\begin{equation*}
\begin{split}
x & = (n - \frac{1}{2}) \: \frac{\lambda}{2} \\\\
x & = (n - \frac{1}{2}) \: \frac{2}{2} \\\\
x & = (n - \frac{1}{2}) \text{ meter} \\\\
n = 2 & \rightarrow x = 1 \frac{1}{2} \text{ meter}
\end{split}
\end{equation*}

A. Persamaan gelombang stationer ujung bebas

Panjang gelombang

\begin{equation*}
\begin{split}
v & = \lambda \:.\: f \\\\
100 & = \lambda \:.\: 50 \\\\
\lambda & = 2 \text{ m}
\end{split}
\end{equation*}

\begin{equation*}
\begin{split}
y & = 2A \cos kx \sin \omega t \\\\
y & = 2A \cos \tfrac{2 \pi}{\lambda} x \sin 2 \pi f t \\\\
y & = 2 \:.\: 20 \cos \tfrac{2 \pi}{2} x \sin 2 \pi \:.\: 50 t \\\\
y & = 40 \cos \pi x \sin 100 \pi t
\end{split}
\end{equation*}

B. Simpangan titik P saat sumber getar sudah bergetar 0,025 detik

\begin{equation*}
\begin{split}
y & = 40 \cos \pi x \sin 100 \pi t \\\\
y & = 40 \cos \pi (0,25) \sin 100 \pi (0,025) \\\\
y & = 40 \cos \tfrac{1}{4}\pi \sin \tfrac{1}{2}\pi \\\\
y & = 40 \:.\: \tfrac{1}{2} \sqrt{2} \:.\: 1 \\\\
y & = 20 \sqrt{2} \text{ cm}
\end{split}
\end{equation*}

C. Kapankah titik P memiliki simpangan ½ dari amplitudo

\begin{equation*}
\begin{split}
y & = 40 \cos \pi x \sin 100 \pi t \\\\
20 & = 40 \cos \pi (0,25) \sin 100 \pi t \\\\
20 & = 40 \:.\: \tfrac{1}{2} \sqrt{2} \sin 100 \pi t \\\\
20 & = 20 \sqrt{2} \sin 100 \pi t \\\\
\frac{1}{\sqrt{2}} & = \sin 100 \pi t \\\\
\frac{1}{2} \sqrt{2} & = \sin 100 \pi t \\\\
\sin \frac{\pi}{4} & = \sin 100 \pi t \\\\
\frac{\pi}{4} & = 100 \pi t \\\\
t & = \frac{1}{400} \text{ detik}
\end{split}
\end{equation*}

A. Amplitudo dan cepat rambat gelombang asal

\begin{equation*}
\begin{split}
y & = 2A \cos kx \sin \omega t \\\\
y & = 0,6 \cos \tfrac{\pi}{6} x \sin 50 \pi t
\end{split}
\end{equation*}

B. Jarak antara simpul dan perut yang berdekatan

Jarak perut dan simpul berdekatan = \(\dfrac {\lambda}{4} = \dfrac {6}{4} = 1,5 \text{ m}\)

C. Kecepatan getar suatu titik pada x = 1,5 m dan t = 9/8 detik

\begin{equation*}
\begin{split}
y & = 0,6 \cos \tfrac{\pi}{6} x \sin 50 \pi t \\\\
v & = \frac{dy}{dt} = 0,6 \:.\: 50 \pi \cos \tfrac{\pi}{6} x \cos 50 \pi t \\\\
v & = 30 \pi \cos \tfrac{\pi}{6} (1,5)  \cos 50 \pi (\tfrac{9}{8}) \\\\
v & = 30 \pi \cos \tfrac{\pi}{4} \cos 56\tfrac{1}{4} \pi \\\\
v & = 30 \pi \cos \tfrac{\pi}{4} \cos \tfrac{1}{4} \pi \\\\
v & = 30 \pi \:.\: \tfrac{1}{2} \sqrt{2} \:.\: \tfrac{1}{2} \sqrt{2} \\\\
v & = 15 \pi \text{ m/s}
\end{split}
\end{equation*}