JEE MAIN - Physics (2003 - No. 42)
The displacement of particle varies according to the relation
$$x=4$$$$\left( {\cos \,\pi t + \sin \,\pi t} \right).$$ The amplitude of the particle is
$$x=4$$$$\left( {\cos \,\pi t + \sin \,\pi t} \right).$$ The amplitude of the particle is
$$-4$$
$$4$$
$$4\sqrt 2 $$
$$8$$
Giải thích
$$x = 4\left( {\cos \pi t + \sin \pi t} \right)$$
$$ = \sqrt 2 \times 4\left( {{{\sin \pi t} \over {\sqrt 2 }} + {{\cos \pi t} \over {\sqrt 2 }}} \right)$$
$$x = 4\sqrt 2 \sin \left( {\pi t + {{45}^ \circ }} \right)$$
$$ = \sqrt 2 \times 4\left( {{{\sin \pi t} \over {\sqrt 2 }} + {{\cos \pi t} \over {\sqrt 2 }}} \right)$$
$$x = 4\sqrt 2 \sin \left( {\pi t + {{45}^ \circ }} \right)$$