«11C» 51. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 51. Hausaufgabe

0.0.1.1 ↑ Blatt

l = 1,\!09\mathrm{m}; \alpha_{\mathrm{max}} = 4,\!0^\circ;$l=1,09\text{m};{\alpha }_{\text{max}}=4,0^{\circ };$

a)

T = 2\pi \sqrt{\dfrac{l}{g}} = \ldots = 2,\!1\mathrm{s};$T=2\pi \sqrt{\frac{l}{g}}=\dots =2,1\text{s};$

A = l \alpha_{\mathrm{max}} = \ldots = 7,\!6\mathrm{cm};$A=l{\alpha }_{\text{max}}=\dots =7,6\text{cm};$

b)

v_{\mathrm{max}} = A \omega = A \dfrac{2\pi}{T} = 0,\!23\frac{\mathrm{m}}{\mathrm{s}};$v_{\text{max}}=A\omega =A\frac{2\pi }{T}=0,23\frac{\text{m}}{\text{s}};$

c)

\alpha = 2,\!0^\circ;$\alpha =2,0^{\circ };$

y = l \alpha = A \cdot \sin \omega{}t = l \alpha_{\mathrm{max}} \sin\!\left[\dfrac{2\pi}{T} t\right]; \Rightarrow \dfrac{\alpha}{\alpha_{\mathrm{max}}} = \sin\!\left[\dfrac{2\pi}{T} t\right];$y=l\alpha =A\cdot sin\omega t=l{\alpha }_{\text{max}}sin\left[\frac{2\pi }{T}t\right];\Rightarrow \frac{\alpha }{{\alpha }_{\text{max}}}=sin\left[\frac{2\pi }{T}t\right];$

⇒ \arcsin\!\left[\dfrac{\alpha}{\alpha_{\mathrm{max}}}\right] = \dfrac{2\pi}{T} t; \Rightarrow t = \dfrac{T}{2\pi} \arcsin\!\left[\dfrac{\alpha}{\alpha_{\mathrm{max}}}\right] = \ldots = 0,\!18\mathrm{s};$arcsin\left[\frac{\alpha }{{\alpha }_{\text{max}}}\right]=\frac{2\pi }{T}t;\Rightarrow t=\frac{T}{2\pi }arcsin\left[\frac{\alpha }{{\alpha }_{\text{max}}}\right]=\dots =0,18\text{s};$

d)

v(t) = A\omega \cdot \cos \omega{}t;$v\left(t\right)=A\omega \cdot cos\omega t;$

v(0,\!18\mathrm{s}) = \ldots = 0,\!20\frac{\mathrm{m}}{\mathrm{s}};$v\left(0,18\text{s}\right)=\dots =0,20\frac{\text{m}}{\text{s}};$