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

Question

Dr. Abubacker · Accepted Answer

(a) 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, $a_{1} = 8\ cm\ \ \ \ and\ {\ \ \ a}_{2} = 6\ cm$
$	herefore$ Amplitude of motion
$A = \sqrt{a_{1}^{2} + a_{2}^{2}}$
= $\sqrt{8^{2} + 6^{2}}$
= $\sqrt{64 + 36} = \ \sqrt{100} = 10\ cm$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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