The displacement equation of a particle is $X = 3 s i n 2 t + 4 c o s 2 t$ . The amplitude and maximum velocity respectively will be

5, 10

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3, 2

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4, 2

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3, 4

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Solution

The correct option is A 5, 10
$X = 3 s i n 2 t + 4 c o s 2 t$
Converting in standard form as,
$X = \sqrt{3^{2} + 4^{2}} s i n (2 t + t a n^{- 1} (4 / 3))$

Comparing with general equation of SHM, $X = A s i n (ω t + ϕ)$
We get $A = 5, V_{m a x} = A ω = 2 \times 5 = 10 m / s$

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