A particle executes S.H.M. of amplitude A alo | vedclass

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

Question

(A) The displacement equation is given by $x(t) = A \sin(\omega t + \delta)$.At $t = 0$,the position is $x = \frac{A}{2}$.Substituting these values: $

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) The displacement equation is given by $x(t) = A \sin(\omega t + \delta)$.At $t = 0$,the position is $x = \frac{A}{2}$.Substituting these values: $\frac{A}{2} = A \sin(\omega(0) + \delta) \Rightarrow \sin \delta = \frac{1}{2}$.This gives two possible values for $\delta$ in the range $[0, 2\pi)$: $\delta = \frac{\pi}{6}$ or $\delta = \frac{5\pi}{6}$.The velocity of the particle is $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \delta)$.At $t = 0$,$v(0) = A\omega \cos \delta$.Since the particle moves along the positive $x$-axis,the velocity must be positive $(v > 0)$,which implies $\cos \delta > 0$.For $\delta = \frac{\pi}{6}$,$\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} > 0$ (Valid).For $\delta = \frac{5\pi}{6}$,$\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} Therefore,the correct value is $\delta = \frac{\pi}{6}$.

$A$ particle executes $S.H.M.$ of amplitude $A$ along the $x$-axis. At $t = 0$,the position of the particle is $x = \frac{A}{2}$ and it moves along the positive $x$-axis. If the displacement of the particle in time $t$ is $x = A \sin (\omega t + \delta)$,then the value of $\delta$ will be:

Similar Questions

$A$ particle starts from a point $P$ at a distance of $A/2$ from the mean position $O$ and travels towards the left as shown in the figure. If the time period of $SHM$ executed about $O$ is $T$ and amplitude is $A$,then the equation of motion of the particle is:

$A$ small mass executes linear $SHM$ about $O$ with amplitude $a$ and period $T.$ Its displacement from $O$ at time $T/8$ after passing through $O$ is:

$A$ particle is executing simple harmonic motion. If the minimum time taken by the particle to move from extreme position to half of the amplitude is $t_1$,and the minimum time taken by the particle to move from mean position to half of the amplitude is $t_2$,then

The displacement versus time curve for a particle executing $SHM$ is shown in the figure. Identify the points marked at which $(i)$ the velocity of the oscillator is zero,$(ii)$ the speed of the oscillator is maximum.

What is the phase difference between two simple harmonic motions represented by $x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right)$ and $x_{2}=A \cos (\omega t)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module