Question

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

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

Answer 1

The correct option is A $10cm$

A harmonic oscillation of constant amplitude and of single frequency is called simple harmonic oscillation.
Here, $x=8\sin \omega t+6\cos \omega t$
So, $a_1=8cm$ and $a_2=6cm$
$\therefore$ Amplitude of motion
$A=\sqrt{a_1^2+a_2^2}$
$=\sqrt{8^2+6^2}$
$=\sqrt{64+36}=\sqrt{100}$
$=10cm$