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Step-by-Step Detailed Solution
Step 1: Express Kinetic Energy and Total Energy in SHM
In Simple Harmonic Motion (SHM), let the amplitude be $a$, the mass be $m$, and the angular frequency be $\omega$. The total energy (T.E.) of the oscillator is given by:
$$
T.E. = \frac{1}{2} m a^2 \omega^2.
$$
At any time $t$, the kinetic energy (K.E.) is:
$$
K.E. = \frac{1}{2} m a^2 \omega^2 \cos^2(\omega t).
$$
Step 2: Set Kinetic Energy to 75% of Total Energy
We are told that the kinetic energy is $75\%\ (=0.75)$ of the total energy. Hence:
$$
K.E. = 0.75 \times T.E.
$$
Substituting the expressions:
$$
\frac{1}{2} m a^2 \omega^2 \cos^2(\omega t) = 0.75 \times \frac{1}{2} m a^2 \omega^2.
$$
Simplifying, we get:
$$
\cos^2(\omega t) = 0.75.
$$
Therefore,
$$
\cos(\omega t) = \sqrt{0.75} = \frac{\sqrt{3}}{2}.
$$
The corresponding angle for which $\cos(\theta) = \tfrac{\sqrt{3}}{2}$ is
$$
\theta = \frac{\pi}{6} \quad (\text{or }30^\circ).
$$
Hence,
$$
\omega t = \frac{\pi}{6}.
$$
Step 3: Relate Angular Frequency to Period and Solve for Time
The period of oscillation $T$ is given as $2\,\text{s}.$ Angular frequency $\omega$ is related to $T$ by:
$$
\omega = \frac{2\pi}{T}.
$$
Substituting $T = 2\,\text{s}$:
$$
\omega = \frac{2\pi}{2} = \pi.
$$
From the relation $\omega t = \frac{\pi}{6}$, we have:
$$
\pi \, t = \frac{\pi}{6}.
$$
Therefore,
$$
t = \frac{1}{6} \,\text{s}.
$$
Final Answer
The body’s kinetic energy will be $75\%$ of the total energy after
$$
\boxed{\frac{1}{6} \text{ s}}.
$$
