A particle is executing the motion x = A cos | vedclass

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

Question

(B) The displacement of the particle is given by $x = A \cos (\omega t - 	heta)$.To find the velocity $v$,we differentiate the displacement with resp

Vedclass · Accepted Answer

$A \omega$

Vedclass · Answer

(B) The displacement of the particle is given by $x = A \cos (\omega t - 	heta)$.To find the velocity $v$,we differentiate the displacement with respect to time $t$:$v = \frac{dx}{dt} = \frac{d}{dt} [A \cos (\omega t - 	heta)]$$v = -A \omega \sin (\omega t - 	heta)$The velocity is maximum when the magnitude of the sine function is $1$,i.e.,$|\sin (\omega t - 	heta)| = 1$.Therefore,the maximum velocity is $v_{\max} = | -A \omega (1) | = A \omega$.Thus,the maximum velocity of the particle is $A \omega$.

$A$ particle is executing the motion $x = A \cos (\omega t - \theta)$. The maximum velocity of the particle is

Similar Questions

The velocity of a particle executing $S.H.M.$ varies with displacement $(x)$ as $4V^2 = 50 - x^2$. The time period of oscillation is $\frac{x}{7}$ seconds. The value of '$x$' is (Take $\pi = \frac{22}{7}$)

$A$ particle performs linear $S.H.M.$ At a particular instant,velocity of the particle is $u$ and acceleration is $\alpha$ while at another instant,velocity is $v$ and acceleration is $\beta$ $(0 < \alpha < \beta)$. The distance between the two positions is

$A$ particle executes simple harmonic motion with amplitude $A$ and period $T$. If it is halfway between the mean position and the extreme position,then its speed at that point is:

$A$ particle is performing simple harmonic motion with amplitude $A$ and angular velocity $\omega$. The ratio of maximum velocity to maximum acceleration is

What is the maximum velocity of a simple harmonic motion with an amplitude of $2 \ m$ and a time period of $2 \ s$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module