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Step-by-Step Solution
Step 1: Identify the key features of SHM at t = 0
In simple harmonic motion (SHM), when the particle is at the mean (equilibrium) position at time
$t = 0$, it has maximum speed. Since kinetic energy (K.E.) depends on the square of velocity,
the kinetic energy at $t = 0$ will be maximum.
Step 2: Expression for velocity and kinetic energy in SHM
If the amplitude of the SHM is $A$ and the angular frequency is $\omega$, then the velocity at any
time $t$ can be written as
$$
v(t) = \omega \sqrt{A^2 - y^2(t)},
$$
where $y(t)$ is the displacement from the mean position at time $t$. The kinetic energy at time
$t$ is given by
$$
\text{K.E.}(t) = \tfrac{1}{2}\, m\, \bigl[v(t)\bigr]^2.
$$
Step 3: Kinetic energy at the extreme position
At the extreme position, the displacement $y = \pm A$, so the velocity is zero there. Thus,
K.E. $= 0$ at $t = \frac{T}{4}, \frac{3T}{4}, \dots$ where $T$ is the time period of oscillation.
Step 4: Nature of the K.E.–time graph
Since the kinetic energy begins at a maximum value at $t = 0$ and becomes zero at $t = \frac{T}{4}$,
it oscillates in a smooth, wave-like manner between K.E.max and 0. By one full time period
$T$, the pattern repeats.
Step 5: Conclusion: Correct graphical option
The correct graph must start at a maximum kinetic energy when $t=0$ and then drop to zero at
$t=\frac{T}{4}$. Hence, the first option ()
depicts this behavior correctly.
