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Step-by-Step Explanation
Step 1: Express the Basic Equations of S.H.M
In simple harmonic motion (S.H.M), the displacement of a particle from its mean position can be written as
$ x(t) = A \cos(\omega t) $ or $ x(t) = A \sin(\omega t), $
where
• $A$ is the amplitude,
• $\omega$ is the angular frequency, and
• $T = \frac{2\pi}{\omega}$ is the time period of oscillation.
Step 2: Relate Force and Acceleration to Displacement
For a particle of mass $m$ in S.H.M:
Acceleration is given by $ a(t) = \frac{d^2x}{dt^2} = -\omega^2 x(t). $
Force is $ F = m \, a(t) = -m \omega^2 x(t). $
Hence, the force and acceleration both depend on the displacement $x$. When $x = 0$, both $a$ and $F$ become zero; when $|x|$ is at maximum (the extremes), $|a|$ (and thus $|F|$) is at maximum.
Step 3: Verify Statement (A) – "The force is zero at t = 3T/4"
• The force $F$ is zero when the displacement $x(t) = 0$.
• For $x(t) = A \cos(\omega t)$, $x(t) = 0$ at $ \omega t = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots $, which correspond to $t = \frac{T}{4}, \frac{3T}{4}, \ldots $
• Thus, at $t = \frac{3T}{4}$, $x = 0 \implies F = 0.$
Therefore, statement (A) is correct.
Step 4: Verify Statement (B) – "The acceleration is maximum at t = T"
• The acceleration $ a(t) = -\omega^2 x(t). $
• The magnitude of acceleration becomes maximum when $|x(t)|$ is maximum, i.e., at the extremes $|x| = A.$
• At $t = T$, $\omega t = \omega \times T = \omega \times \frac{2\pi}{\omega} = 2\pi.$
• Then $ x(T) = A \cos(2\pi) = A,$ which is the extreme position. Hence, $|a| = \omega^2 A$, which is maximum in magnitude.
Therefore, statement (B) is correct.
Step 5: Verify Statement (C) – "The speed is maximum at t = T/4"
• The velocity in S.H.M is $ v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t) \ $ (assuming $x(t) = A \cos(\omega t)$).
• The magnitude of velocity $|v(t)|$ is maximum when $|\sin(\omega t)| = 1.$
• This happens at $ \omega t = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots $, i.e., $ t = \frac{T}{4}, \frac{3T}{4}, \ldots $
• Thus, at $t = \frac{T}{4}$, the speed reaches its maximum value $A\omega.$
Therefore, statement (C) is correct.
Step 6: Check Statement (D) – "The P.E. is equal to K.E. at t = T/2"
• In S.H.M, the potential energy (P.E.) is $ \frac{1}{2} k x^2 $ and the kinetic energy (K.E.) is $ \frac{1}{2} k (A^2 - x^2), $ where $k = m \omega^2.$
• For P.E. = K.E., we need
$ \frac{1}{2} k x^2 = \frac{1}{2} k (A^2 - x^2) \implies x^2 = \frac{A^2}{2}. $
• Thus $ |x| = \frac{A}{\sqrt{2}}.$
• This position occurs at times $t = \frac{T}{8}, \frac{3T}{8}, \ldots$ not at $t = \frac{T}{2}.$
• At $t = \frac{T}{2}$, the particle is at $x = -A$ (an extreme), so the P.E. is maximum and K.E. is zero, contradicting statement (D).
Hence, statement (D) is not correct.
Conclusion
The correct statements are (A), (B), and (C), and therefore the correct answer is indeed
(A) The force is zero at $t = \frac{3T}{4}$.
(B) The acceleration is maximum at $t = T.$
(C) The speed is maximum at $t = \frac{T}{4}$.
