The figure shows a graph between velocity (v) | vedclass

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(D) From the graph,we can see that the maximum velocity $V_{\max} = 10 \, cm/s$ and the amplitude $A = 2.5 \, cm$.We know that $V_{\max} = A\omega$.Su

Vedclass · Accepted Answer

All of the above.

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(D) From the graph,we can see that the maximum velocity $V_{\max} = 10 \, cm/s$ and the amplitude $A = 2.5 \, cm$.We know that $V_{\max} = A\omega$.Substituting the values: $10 = 2.5 \omega \Rightarrow \omega = 4 \, s^{-1}$.The time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{3.14}{2} = 1.57 \, s$. Thus,option $(A)$ is correct.The maximum acceleration $a_{\max} = A\omega^2 = 2.5 	imes 4^2 = 2.5 	imes 16 = 40 \, cm/s^2$. Thus,option $(B)$ is correct.The velocity at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.For $x = 1 \, cm$,$v = 4 \sqrt{2.5^2 - 1^2} = 4 \sqrt{6.25 - 1} = 4 \sqrt{5.25} = 4 \sqrt{\frac{525}{100}} = 4 	imes \frac{\sqrt{21 	imes 25}}{10} = 4 	imes \frac{5\sqrt{21}}{10} = 2\sqrt{21} \, cm/s$. Thus,option $(C)$ is correct.Since all options are correct,the correct answer is $(D)$.

Column $I$	Column $II$
$(A)$ The object moves on the $x$-axis under a conservative force such that its speed $v = c_1 \sqrt{c_2 - x^2}$,where $c_1, c_2 > 0$.	$(p)$ The object executes simple harmonic motion.
$(B)$ The object moves on the $x$-axis such that its velocity $v = -kx$,where $k > 0$.	$(q)$ The object does not change its direction.
$(C)$ An object is attached to a spring in an elevator accelerating upwards with constant acceleration $a$. The motion is observed from the elevator.	$(r)$ The kinetic energy of the object keeps on decreasing.
$(D)$ The object is projected vertically upwards with speed $2 \sqrt{GM_e / R_e}$.	$(s)$ The object can change its direction only once.

The figure shows a graph between velocity $(v)$ and displacement $(x)$ from the mean position of a particle performing $SHM$.

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