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

(A) $10 \mathrm{~cm}, \frac{2}{3} \mathrm{~s}$

(B) $10 \mathrm{~cm}, \frac{3}{2} \mathrm{~s}$

(C) $5 \mathrm{~cm}, \frac{3}{2} \mathrm{~s}$

(D) $5 \mathrm{~cm}, \frac{2}{3} \mathrm{~s}$

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Elasticity & SHM

A simple harmonic motion is represented by :

y = 5 (sin 3 π t + 3 cos 3 π t) cm

The amplitude and time period of the motion are :

Answer: (A)

Write

y = 5 (sin 3 π t + 3 cos 3 π t)

in the single-sinusoid form

A sin (ω t + ϕ)

. The amplitude is

A = 5 1^{2} + (3)^{2} = 54 = 10

cm, and with

ω = 3 π

rad/s the period is

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

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option