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(C) Given,amplitude $A = 2 \,m$. Maximum acceleration $a_{max} = A \omega^2$ and maximum velocity $v_{max} = A \omega$. According to the problem,$|A \

Vedclass · Accepted Answer

$\frac{22}{7} \,s, 2 \sqrt{3} \,ms^{-1}$

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(C) Given,amplitude $A = 2 \,m$. Maximum acceleration $a_{max} = A \omega^2$ and maximum velocity $v_{max} = A \omega$. According to the problem,$|A \omega^2| - |A \omega| = 4$. Substituting $A = 2$: $2 \omega^2 - 2 \omega = 4$ $\omega^2 - \omega - 2 = 0$ $(\omega - 2)(\omega + 1) = 0$. Since $\omega > 0$,we get $\omega = 2 \,rad/s$. Time period $T = \frac{2 \pi}{\omega} = \frac{2 \pi}{2} = \pi = \frac{22}{7} \,s$. Velocity at displacement $y = 1 \,m$ is given by $v = \omega \sqrt{A^2 - y^2}$. $v = 2 \sqrt{2^2 - 1^2} = 2 \sqrt{4 - 1} = 2 \sqrt{3} \,ms^{-1}$.

$A$ particle is executing simple harmonic motion with an amplitude of $2 \,m$. The difference in the magnitudes of its maximum acceleration and maximum velocity is $4$. The time-period of its oscillation and its velocity when it is $1 \,m$ away from the mean position are respectively:

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