A simple harmonic motion is represented by y | vedclass

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Question

(A) Given equation: $y = 5(\sin 3\pi t + \sqrt{3} \cos 3\pi t) \ cm$.We can rewrite this in the form $y = A \sin(\omega t + \phi)$.Multiply and divide

Vedclass · Accepted Answer

$10 \ cm, \frac{2}{3} \ s$

Vedclass · Answer

(A) Given equation: $y = 5(\sin 3\pi t + \sqrt{3} \cos 3\pi t) \ cm$.We can rewrite this in the form $y = A \sin(\omega t + \phi)$.Multiply and divide by $2$: $y = 5 	imes 2 \left( \frac{1}{2} \sin 3\pi t + \frac{\sqrt{3}}{2} \cos 3\pi t ight)$.Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we set $\cos \phi = 1/2$ and $\sin \phi = \sqrt{3}/2$,which gives $\phi = \pi/3$.So,$y = 10 \sin(3\pi t + \pi/3) \ cm$.The amplitude $A = 10 \ cm$.The angular frequency $\omega = 3\pi \ rad/s$.The time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{3\pi} = \frac{2}{3} \ s$.

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

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