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

• A. 10 cm
• B. 2 cm
• C. 14 cm
• D. 3.5 cm

Answer 1

Answer: (A)

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$