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

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

Question

(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.

Vedclass · Answer

(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 $(D)$.

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

Similar Questions

$A$ graph of the square of the velocity against the square of the acceleration of a given simple harmonic motion is

$A$ particle executing simple harmonic motion has an amplitude of $6\, cm$. Its acceleration at a distance of $2 \,cm$ from the mean position is $8\, cm/s^2$. The maximum speed of the particle is ... $cm/s$.

$A$ horizontal platform with an object placed on it is executing $S.H.M.$ in the vertical direction. The amplitude of oscillation is $3.92 \times 10^{-3} \, m$. What must be the least period of these oscillations,so that the object is not detached from the platform (in $, s$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module