NEET Equation of SHM MCQs & PYQs

The expression \(x = 4sin^2\omega t + 2cos^2\omega t + 2sin\omega t cos\omega t\) simplifies to \(x = 3 + sin(2\omega t) - cos(2\omega t)\). The oscillatory part is \(sin(2\omega t) - cos(2\omega t)\). Assertion (A) is true, its amplitude is \(\sqrt{1^2 + (-1)^2} = \sqrt{2}\). Reason (R) is true, the angular frequency is \(2\omega\). However, the angular frequency does not explain the specific amplitude value.

Assertion (A) is true. If displacement \(x = Asin(\omega t)\), then acceleration \(a = -A\omega^2sin(\omega t) = A\omega^2sin(\omega t + \pi)\). Reason (R) is true. Velocity \(v = A\omega cos(\omega t) = A\omega sin(\omega t + \pi/2)\). Both statements are true, but the phase relationship of velocity with displacement does not explain the phase relationship of acceleration with displacement directly; they are separate facts of SHM.

For a harmonic oscillator, velocity \( v \) and displacement \( x \) are related by \( v = \omega \sqrt{A^2 - x^2} \). Squaring this gives \( v^2 = \omega^2 (A^2 - x^2) \), which is an equation of an ellipse, not a parabola. So (A) is false. Velocity does not change uniformly with displacement, hence (R) is also false. Thus, both A and R are false.

Periodic functions like sine and cosine repeat their values over a fixed period. Reason (R) defines periodicity, which directly explains Assertion (A).
Thus, both are true, and R explains A.

In SHM, velocity is max at equilibrium (where displacement is zero), and acceleration is min (zero) at equilibrium. So A is true.
Displacement `\(x = A\sin(\omega t)\)` and velocity `\(v = A\omega\cos(\omega t)\)` differ in phase by \(\frac{\pi}{2}\). So R is true.
However, R explains phase relation, not why maximum velocity occurs at minimum acceleration. Hence, R does not explain A.

Vibration of polyatomic molecules involves multiple normal modes, each with a different frequency. The total vibration is a superposition of these individual SHMs.
This complex, multi-frequency nature means the overall motion is not a single SHM. Both A and R are true, and R explains A.

Comparing with the standard equation of SHM, \(\frac{d^2y}{dt^2} + \omega^2 y = 0\), we get \(\omega = \sqrt{K}\). Therefore, the time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{K}}\).

The time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/2} = 4\text{ s}\). In \(t = 3\text{ s}\) (which is \(\frac{3T}{4}\)), the particle completes three quarters of an oscillation, covering a total distance of \(3a\).

Since the two perpendicular components have a phase difference of \(\frac{\pi}{2}\), the net amplitude is \(A = \sqrt{\alpha^2 + \beta^2}\).

The maximum acceleration is given by \(a_{\text{max}} = \omega^2 A = (2\pi f)^2 A\). Substituting \(f = 4\text{ Hz}\) and \(A = 0.15\text{ m}\) gives \(a_{\text{max}} = 4\pi^2 (16)(0.15) \approx 94.65\text{ m/s}^2\).