Begini Pertanyaanya.....
Soal. Sebuah benda bermassa 200 gram tergantung pada sebuah pegas dan bergetar mengikuti gerak harmonik sederhana dengan persamaan $y=2\sin \left( 20\pi t+\frac{\pi}{3} \right)\text{ cm}$. Tentukan besar:
- Amplitudo dan frekuensi getaran;
- Simpangan saat detik ke 5;
- Kecepatan getar saat detik ke 5;
- Percepatan saat detik ke 5;
- Kecepatan maksimum;
- Energi kinetik saat simpangan 1 cm;
- Energi mekanik;
- Simpangan saat Ek = 3Ep.
1. Besar amplitudo, frekuensi, perioda, dan sudut fase awal;\begin{align*} y&=2\sin \left( 20\pi t+\frac{\pi}{3} \right)\text{ cm}\\y&=A\sin (\omega t\pm {{\theta }_{0}})\\& \textrm{ diperoleh:}\\&\textrm{Amplitudo }2 \, cm\\&\textrm{Frekuensi: } \omega =20\pi\\&2\pi f=20\pi\\&f=10Hz \end{align*}
2. Simpangan saat detik ke 5 \[\begin{align*} y&=2\sin \left( 20\pi t+\frac{\pi}{3} \right)|_{t=5}\\y&=2\sin \left( 20\pi (5)+\frac{\pi}{3} \right)\\y&=2\sin \left(\frac{\pi}{3} \right)=2\sin 60^o\\y&=2\left ( \frac{\sqrt{3}}{2} \right ) =\sqrt{3}\, \, cm\end{align*}\]
3. Kecepatan getar pada detik ke 5; \[\begin{align*} v&=\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{\mathrm{d} (2\sin \left( 20\pi t+\frac{\pi}{3} \right))}{\mathrm{d} t} \\v&=2\left( 20\pi \right)\cos \left( 20\pi t+\frac{\pi}{3} \right) |_{t=5} \\v&= 40\pi \cos \left( 20\pi (5)+\frac{\pi}{3} \right) \\v&=40\pi \cos \left(\frac{\pi}{3} \right) \\v&=40\pi \cos 60^{o}\\v&=40\pi (0,5)=20\pi cm.s^{-1}\end{align*}\]Kita bisa juga langsung menggunakan rumus percepatan: \[\begin{align*} v&=A{\omega }\cos \left( \omega t+{{\theta }_{0}} \right)\\ v&=2( 20\pi )\cos \left( 20\pi t+\frac{\pi}{3} \right)|_{t=5}\\ v&=40\pi \cos \left( 20\pi (5)+\frac{\pi}{3} \right)\\ v&=40\pi \cos 60^o\\ v&=40\pi (0,5)\\v&=20\pi \: \: cm.s^{-1} \end{align*}\]
4. Besar percepatan saat detik ke 5:\[\begin{align*} a&=\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} \left [40\pi \cos (20\pi t+\frac{\pi}{3} ) \right ]}{\mathrm{d} t} \\a&=-40\pi \left( 20\pi \right)\sin \left( 20\pi t+\frac{\pi}{3} \right) |_{t=5} \\a&= -800\pi ^2 \sin \left( 20\pi (5)+\frac{\pi}{3} \right) \\a&=-800\pi ^2 \sin \left(\frac{\pi}{3} \right) \\a&=-800\pi ^2 \sin 60^{o}\\a&=-800\pi ^2 (0,5\sqrt{3})\\a&=-400\sqrt{3}\pi ^2 cm.s^{-2}\end{align*}\]Kita bisa juga langsung menggunakan rumus percepatan: \[\begin{align*} a&=-A{{\omega }^{2}}\sin \left( \omega t+{{\theta }_{0}} \right)\\ a&=-2{{\left( 20\pi \right)}^{2}}\sin \left( 20\pi t+\frac{\pi}{3} \right)|_{t=5}\\ a&=-800\pi^{2}\sin \left( 20\pi (5)+\frac{\pi}{3} \right)\\ a&=-800\pi ^2 \sin 60^{o}\\a&=-800\pi ^2 (0,5\sqrt{3})\\a&=-400\sqrt{3}\pi ^2 cm.s^{-2}\end{align*}\]
5. Kecepatan maksimum:\[\begin{align*} v_{maks}&=\omega A\\ v_{maks}&=20\pi (2)\\v_{maks}&=40\pi \: cm.s^{-1}\\v_{maks}&=0,4\pi \: m.s^{-1}\end{align*}\]
6. Energi kinetik dan energi potensial saat simpangan 1 cm: \[\begin{align*}E_k&=\frac{1}{2}mv^2\\E_k&=\frac{1}{2}m\left [ \omega \sqrt{A^2-y^2} \right ]^2\\ E_k&=\frac{1}{2}m\omega ^2\left [ A^2-y^2\right ]\\E_k&=\frac{1}{2}(200)(20\pi ) ^2\left [2^2-1^2 \right ]\\E_k&=100(400\pi ^2\left [3\right ]\\E_k&=120000\pi ^2\: erg|_{ambil \: \pi^2=10}\\E_k&\approx 0,12\times 10^{-7}erg\\E_k&\approx 0,12\, J\end{align*}\]
7. Energi mekanik: \[\begin{align*}E_m&=\frac{1}{2}kA^2\\E_m&=\frac{1}{2}m\omega ^{2}A^2 \\E_k&=\frac{1}{2}(200)(20\pi ) ^2(2)^2\\E_k&=100(400\pi ^2\left [4\right ]\\E_k&= 160000\pi ^2\: erg|_{ambil \: \pi^2=10}\\E_k&\approx 0,16\times 10^{-7}erg\\E_k& \approx 0,16\, J\end{align*}\]
8. Simpangan saat Ek = 3Ep
Untuk gerak harmonik sederhana, jika $E_k=nE_p$ maka:\[\begin{align*}y&=\frac{A}{\sqrt{n+1}}\\y&=\frac{2}{\sqrt{3+1}}\\y&=1\: cm\end{align*}\]Terimakasih..
