Physics Question on Oscillations

Question

The displacement of a particle performing simple harmonic motion is given by, x=8sin $\omega t+6cos \omega t$ , where distance is in cm and time is in second. The amplitude of motion is

Answer 1

Solution

Here $x=8 sin \omega t+6 cos \omega t$
so, $\, \, \, \, \, \, \, a_1 =8cm \, and \, a_2 =6cm$
$\therefore \, \, \, \, \,$ Amplitude of motion
$\, \, \, \, \, \, \, \, \, A =\sqrt{a_1^2 +a_2^2}$
$\, \, \, \, \, \, \, \, \, =\sqrt{8^2}+6^2$
$\, \, \, \, \, \, \, \, \, \, \, =\sqrt{64+36}=\sqrt{100}$
$\, \, \, \, \, \, \, \, \, \, \, =10 cm$