A particle executes S.H.M. according to the e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given equation for displacement is $x = 10 \cos \left(2 \pi t + \frac{\pi}{2}\right)$.The velocity $v$ is the derivative of displacement with

Vedclass · Accepted Answer

$31.4$

Vedclass · Answer

(D) The given equation for displacement is $x = 10 \cos \left(2 \pi t + \frac{\pi}{2}\right)$.The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt}$.$v = \frac{d}{dt} \left[10 \cos \left(2 \pi t + \frac{\pi}{2}\right)\right] = -10 \cdot 2 \pi \sin \left(2 \pi t + \frac{\pi}{2}\right)$.$v = -20 \pi \sin \left(2 \pi t + \frac{\pi}{2}\right)$.At $t = \frac{1}{6} \, s$,the velocity is:$v = -20 \pi \sin \left(2 \pi \cdot \frac{1}{6} + \frac{\pi}{2}\right) = -20 \pi \sin \left(\frac{\pi}{3} + \frac{\pi}{2}\right)$.$v = -20 \pi \sin \left(\frac{5\pi}{6}\right)$.Since $\sin \left(\frac{5\pi}{6}\right) = \sin \left(150^\circ\right) = \frac{1}{2}$,we have:$v = -20 \pi \cdot \frac{1}{2} = -10 \pi$.The magnitude of velocity is $|v| = 10 \pi \approx 10 \cdot 3.14 = 31.4 \, cm/s$.

$A$ particle executes $S.H.M.$ according to the equation $x=10 \cos \left[2 \pi t+\frac{\pi}{2}\right] \, cm$,where $t$ is in seconds. The magnitude of the velocity of the particle at $t=\frac{1}{6} \, s$ will be .............. $cm/s$.

Similar Questions

The maximum velocity and the maximum acceleration of a body moving in a simple harmonic oscillator are $2 \ m/s$ and $4 \ m/s^2$,respectively. Then the angular velocity will be ..... $rad/s$.

$A$ 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 is equal to the magnitude of its acceleration. Then,its periodic time in seconds is

The graph in the figure represents:

$A$ mass attached to a spring performs $S.H.M.$ whose displacement is $x = 3 \times 10^{-3} \cos(2 \pi t) \text{ m}$. The time taken to obtain maximum speed for the first time is

Vedclass Test Series

Exam Paper Generator

Online Exam Module