Question 211: Oscillations

Question

The displacement of particle in S.H.M. is $\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}+\pi / 6)$. Its speed will be maximum at time $\left(\sin 90^{\circ}=1\right)$

Accepted Answer

Displacement:
$ x = A \cos(\omega t + 	frac{\pi}{6}) $
We want the time when
speed is maximum
.
Step 1: Velocity in SHM
$ v = \frac{dx}{dt} = -A \omega \sin(\omega t + 	frac{\pi}{6}) $
So:
$ v^2 = A^2 \omega^2 \sin^2(\omega t + 	frac{\pi}{6}) $
Maximum speed:
$ v_{\max} = A \omega $
occurs when
$ |\sin(\omega t + 	frac{\pi}{6})| = 1 $
i.e.
$ \sin(\omega t + 	frac{\pi}{6}) = \pm 1 $
Step 2: Condition for maximum
So:
$ \omega t + 	frac{\pi}{6} = \frac{\pi}{2}, \; \frac{3\pi}{2}, \; \dots $
Choose the smallest positive time. First solution:
$ \omega t + 	frac{\pi}{6} = \frac{\pi}{2} $
$ \omega t = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} $
$ t = \frac{\pi}{3\omega} $
This is the
first time
when speed is maximum.
Step 3: Match with options
Correct option:
A) $	frac{\pi}{3\omega}$
Ã¢ÂÂ Final Answer:
$ \boxed{	frac{\pi}{3\omega}} $

Answer

$\frac{\pi}{3 \omega} \mathrm{~s}$

Answer

$\frac{\pi}{2 \omega} \mathrm{~s}$

Answer

$\frac{\pi}{\omega} \mathrm{s}$

Answer

$\frac{\pi}{4 \omega} \mathrm{~s}$

Oscillations

The displacement of particle in S.H.M. is \mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}+\pi / 6). Its speed will be maximum at time \left(\sin 90^{\circ}=1\right)

Choose the correct answer: