Time Required for Particle in SHM to Move Between Positions

To find the time required for a particle executing Simple Harmonic Motion (SHM) to move from $x = A$ to $x = \frac{A}{\sqrt{2}}$, we can set up the equation of motion.

Since the particle starts its motion from the extreme position $x = A$ at time $t = 0$, its displacement as a function of time is given by: $x(t) = A \cos(\omega t)$

where:

• $A$ is the amplitude of oscillation,
• $\omega = \frac{2\pi}{T}$ is the angular frequency,
• $T = 5\text{ sec}$ is the time period.

We need to find the time $t$ when $x = \frac{A}{\sqrt{2}}$. Substituting this value into the displacement equation: $\frac{A}{\sqrt{2}} = A \cos(\omega t)$

Dividing both sides by $A$: $\cos(\omega t) = \frac{1}{\sqrt{2}}$

Taking the inverse cosine for the first passage from $x = A$ to $x = \frac{A}{\sqrt{2}}$: $\omega t = \frac{\pi}{4}$

Now, substitute $\omega = \frac{2\pi}{T}$: $\left(\frac{2\pi}{T}\right) t = \frac{\pi}{4}$

Solving for $t$: $t = \frac{T}{8}$

Given that the time period $T = 5\text{ sec}$: $t = \frac{5}{8}\text{ sec}$

Hence, the correct option is C (or $5/8$).