Displacement-time equation of a particle executing SHM is, $x = 4\sin{\omega t + 3\ sin(\omega t + \pi/3}).$ Here $x$ is in centimeter and t in second. The amplitude of oscillation of the particle is approximately

Question

Dr. Abubacker · Accepted Answer

(b) $x = 3\sin{\omega t} + 4\sin{(\omega t + \pi/3)}$
Comparing it with the equations
$x = r_{1}\sin{\omega t + r_{2}\sin{(\omega t + \phi)}}$
We have, $r_{1} = 3$ cm, $r_{2} = 4$cm and $\phi = \pi/3$
The amplitude of combination is
$r = \sqrt{r_{1}^{2} + r_{2}^{2} + 2r_{1}r_{2}\cos\phi}$
$= \sqrt{3^{2} + 4^{2} + 2 	imes 3 	imes 4 	imes \cos{\pi/3}}$
$= \sqrt{37} = 6$ cm

Q. Displacement-time equation of a particle executing SHM is, $x = 4\sin{\omega t + 3\ sin(\omega t + \pi/3}).$ Here $x$ is in centimeter and t in second. The amplitude of oscillation of the particle is approximately

Explanation

Reviewed by Dr. Abubacker

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