Question 130 of 194

A simple harmonic motion is represented by:
$y=5\left(\sin 3\pi t+\sqrt{3}\cos 3\pi t\right)cm$
The amplitude and time period of the motion are:

A$5 cm, \dfrac{2}{3} s$B$10 cm, \dfrac{2}{3} s$C$5 cm, \dfrac{3}{2} s$D$10 cm, \dfrac{3}{2} s$

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Answer: (B) $10 cm, \dfrac{2}{3} s$

$y=5\sin \left(3\pi t\right)+5\sqrt{3}\cos \left(3\pi t\right)$



${A}^{2}={5}^{2}+{\left(5\sqrt{3}\right)}^{2}={10}^{2}$

$A=10cm$

$\omega =3\pi$

$T=\dfrac{2\pi }{\omega }=\dfrac{2}{3}s$

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