If $x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m}$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
NEET 2024, Medium
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$y_1 =10 \sin \left(\omega t+\frac{\pi}{3}\right) cm$
$y_2 =5[\sin (\omega t)+\sqrt{3} \cos \omega t] \;cm$ respectively.
The amplitude of the resultant wave is $.............cm$.
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