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(C) The general equation for simple harmonic motion $(SHM)$ is given by $x = A \cos(\omega t)$, where $A$ is the amplitude and $\omega$ is the angular

Vedclass · Accepted Answer

$5/8$

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(C) The general equation for simple harmonic motion $(SHM)$ is given by $x = A \cos(\omega t)$, where $A$ is the amplitude and $\omega$ is the angular frequency.At the starting position $x = A$, we have $A = A \cos(\omega t_1)$, which implies $\cos(\omega t_1) = 1$, so $t_1 = 0$.At the position $x = A/\sqrt{2}$, we have $A/\sqrt{2} = A \cos(\omega t_2)$, which simplifies to $\cos(\omega t_2) = 1/\sqrt{2}$.This gives $\omega t_2 = \pi/4$.Given that the time period $T = 5 	ext{ sec}$, the angular frequency is $\omega = 2\pi/T = 2\pi/5 	ext{ rad/sec}$.Substituting the value of $\omega$, we get $(2\pi/5) t_2 = \pi/4$.Solving for $t_2$, we find $t_2 = (5 	imes \pi) / (4 	imes 2\pi) = 5/8 	ext{ sec}$.The time required to move from $x = A$ to $x = A/\sqrt{2}$ is $\Delta t = t_2 - t_1 = 5/8 - 0 = 5/8 	ext{ sec}$.

$A$ particle is executing simple harmonic motion. Its amplitude is $A$ and time period is $5 \text{ sec}$. The time required by it to move from $x = A$ to $x = A/\sqrt{2}$ is . . . . . . sec.

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