JEE PYQ: Oscillations Question 35
Question 35 - 2019 (10 Jan 2019 Shift 2)
A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is:
(1) $\frac{4\pi}{3}$
(2) $\frac{3}{8}\pi$
(3) $\frac{8\pi}{3}$
(4) $\frac{7}{3}\pi$
Show Answer
Answer: (3)
Solution
$|v| = |a|$: $\omega\sqrt{25-16} = \omega^2 \times 4$. $3\omega = 4\omega^2$. $\omega = 3/4$. $T = \frac{8\pi}{3}$.
BETA
