JEE PYQ: Oscillations Question 42
Question 42 - 2019 (12 Jan 2019 Shift 1)
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:
(1) 10 cm, $\frac{2}{3}$ s
(2) 10 cm, $\frac{3}{2}$ s
(3) 5 cm, $\frac{3}{2}$ s
(4) 5 cm, $\frac{2}{3}$ s
Show Answer
Answer: (1)
Solution
$y = 10\sin(3\pi t + \frac{\pi}{3})$. Amplitude = 10 cm. $T = \frac{2}{3}$ s.