A simple harmonic motion is represented by $y = 5 (sin 3 π t + 3 cos 3 π t) c m$ The amplitude and time period of the motion are :

Question

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Answer 1

$y = 5 (sin 3 π t + 3 cos 3 π t)$
$y = 5 \times 2 (\frac{1}{2} sin 3 π t + \frac{3}{2} cos 3 π t)$
$y = 10 (cos \frac{π}{3} sin 3 π t - sin \frac{π}{3} cos 3 π t)$
$y = 10 sin (3 π t + \frac{π}{3})$
Comparing it with $y = A sin (ω t + ϕ)$
$\Rightarrow A = 10$ i.e., amplitude is $10$ cm
$\Rightarrow T = \frac{2 π}{ω} = \frac{2 π}{3 π} = \frac{2}{3}$ sec. i.e, time period is $\frac{2}{3}$ sec.

Answer 2

y = 5 (sin 3 π t + 3 cos 3 π t)

y = 5 \times 2 (\frac{1}{2} sin 3 π t + \frac{3}{2} cos 3 π t)

y = 10 (cos \frac{π}{3} sin 3 π t - sin \frac{π}{3} cos 3 π t)

y = 10 sin (3 π t + \frac{π}{3})

Comparing it with

y = A sin (ω t + ϕ)

\Rightarrow A = 10

i.e., amplitude is

10

cm

\Rightarrow T = \frac{2 π}{ω} = \frac{2 π}{3 π} = \frac{2}{3}

sec. i.e, time period is

\frac{2}{3}

sec.