The graph in the figure represents:

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

Question

(A) For a simple harmonic oscillator,the displacement is given by $x(t) = A \sin(\omega t + \phi)$.The velocity is given by $v(t) = \frac{dx}{dt} = A\

Vedclass · Accepted Answer

Motion of a simple pendulum starting from the mean position.

Vedclass · Answer

(A) For a simple harmonic oscillator,the displacement is given by $x(t) = A \sin(\omega t + \phi)$.The velocity is given by $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.From the given graph,at $t = 0$,the velocity $v(0)$ is maximum (positive peak).If the pendulum starts from the mean position $(x=0)$,the velocity is maximum at $t=0$,which matches the graph.If the pendulum starts from the extreme position $(x=A)$,the velocity is $0$ at $t=0$,which does not match the graph.Therefore,the graph represents the motion of a simple pendulum starting from the mean position.

Similar Questions

What is the maximum velocity of a simple harmonic motion with an amplitude of $3 \ cm$ and a time period of $6 \ s$?

$A$ particle executes $S.H.M.$ with amplitude $a$ and time period $T$. The displacement of the particle when its speed is half of its maximum speed is $\frac{\sqrt{x} a}{2}$. The value of $x$ is $\ldots \ldots \ldots$

When a particle executes simple harmonic motion,the nature of the graph of velocity as a function of displacement will be:

Maximum speed of a particle in simple harmonic motion is $v_{max}.$ Then average speed of a particle in one complete oscillation is equal to

$A$ particle executing simple harmonic motion has a maximum speed of $40 \,ms^{-1}$ and maximum acceleration of $60 \,ms^{-2}$. The period of oscillation is

Vedclass Test Series

Exam Paper Generator

Online Exam Module