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: (a) 10 cm, 3/2 s (b) 5 cm, 2/3 s (c) 5 cm, 3/2 s (d) 10 cm, 2/3 s

Answer: (d). For asin θ + bcos θ, amplitude = √(a² + b²). Amplitude = 5√(1² + (√(3))²) = 5 × 2 = 10 cm. ω = 3π, so T = (2π)/(3π) = 2/3 s. So the answer is (d) 10 cm, 2/3 s.

A simple harmonic motion is represented by: y = 5 (sin 3 π t + √(3)… | JEE Main 2019 (12 Jan, Shift 2)

Q58JEE Main 2019 · 12 JAN · SHIFT 2Easy

Periodic Motion & SHM Basics

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:

Show answer and solution

Answer: (d)

For $a\sin\theta + b\cos\theta$ , amplitude $= \sqrt{a^2 + b^2}$ .
Amplitude $= 5\sqrt{1^2 + (\sqrt{3})^2} = 5 \times 2 = 10\ \text{cm}$ .
$\omega = 3\pi$ , so $T = \frac{2\pi}{3\pi} = \frac{2}{3}\ \text{s}$ .
So the answer is (d) 10 cm, $\frac{2}{3}$ s.

Report a mistake

Periodic Motion & SHM Basics question from JEE Main 2019 (12 Jan, Shift 2)

More questions on Periodic Motion & SHM Basics