For a particle executing S.H.M.,where x is th | vedclass

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(D) For a particle in $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t

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all of the above

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(D) For a particle in $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.The acceleration is $a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) = -\omega^2 x$.$1$. For the $v-x$ graph: Since $x = A \sin(\omega t + \phi)$ and $v = A \omega \cos(\omega t + \phi)$,we have $(x/A)^2 + (v/A\omega)^2 = 1$,which represents an ellipse.$2$. For the $a-x$ graph: Since $a = -\omega^2 x$,this is a linear equation of the form $y = mx$,which represents a straight line passing through the origin.$3$. For the $a-v$ graph: Since $v = A \omega \cos(\omega t + \phi)$ and $a = -A \omega^2 \sin(\omega t + \phi)$,we have $(v/A\omega)^2 + (a/A\omega^2)^2 = 1$,which represents an ellipse.Therefore,all the given statements are correct.

Column-$I$	Column-$II$
$(A)$ Velocity-displacement graph $(\omega \neq 1)$	$(i)$ Straight line
$(B)$ Acceleration-displacement graph	$(ii)$ Sinusoidal
$(C)$ Acceleration-time graph	$(iii)$ Circle
$(D)$ Acceleration-velocity graph $(\omega \neq 1)$	$(iv)$ Ellipse

For a particle executing $S.H.M.$,where $x$ is the displacement from the equilibrium position,$v$ is the velocity at any instant,and $a$ is the acceleration at any instant,then:

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The displacement-time graph of a particle executing $SHM$ is shown. Which of the following statements is wrong?

Two particles are in $SHM$ in a straight line. Amplitude $A$ and time period $T$ of both the particles are equal. At time $t=0$,one particle is at displacement $y_1 = +A$ and the other at $y_2 = -A/2$,and they are approaching towards each other. After what time do they cross each other?

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