NEET Oscillation MCQs & PYQs

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.

The total energy of an SHM is `\(E = \frac{1}{2}kA^2\)`. If amplitude `\(A\)` is doubled, energy becomes `\(E' = \frac{1}{2}k(2A)^2 = 4E\)`. So A is false.
Reason R states energy is directly proportional to amplitude, which is also false (it's proportional to `\(A^2\)`). Both are false.

For an ideal SHM, mechanical energy is conserved (A is true). Kinetic energy `\(KE = \frac{1}{2}m\omega^2A^2\cos^2(\omega t)\)` and potential energy `\(PE = \frac{1}{2}k A^2\sin^2(\omega t)\)` vary with `\(2\omega\)`, double the SHM frequency (R is true).
However, R describes the variation of KE/PE, not the reason for conservation of total mechanical energy. Thus, R does not explain A.

The effective acceleration due to gravity is \(g' = g\left(1 - \frac{rho_l}{\rho_b}\right) = g\left(1 - \frac{1}{4}\right) = \frac{3}{4}g\). The new time period is \(T' = T\sqrt{\frac{g}{g'}} = \sqrt{3} \times \sqrt{\frac{4}{3}} = 2 \text{s}\).

Using standard SHM formula \(x = A sin(\omega t + phi)\), where \(A = 0.1 \text{m}\), \(\omega = 2\pi f = 2\pi(25) = 50\pi \text{rad/s}\), and \(\phi = \frac{\pi}{4}\). Substituting gives \(x = 0.1 sin \left(50\pi t + \frac{\pi}{4}\right)\).

Total mechanical energy \(E\) in SHM is the sum of kinetic energy \(K\) and potential energy \(U\): \(E = K + U\). Substituting the values, \(8\text{ J} = 10\text{ J} + U\), which gives \(U = -2\text{ J}\) as the potential energy at that instant.

The time period of a simple pendulum is \(T = 2\pi \sqrt{\frac{l}{g}}\). For a second's pendulum, \(T = 2\text{ s}\). Thus, \(2 = 2\pi \sqrt{\frac{l}{10}} ⇒ 1 = \pi^2 \frac{l}{10}\). Since \(\pi^2 \approx 10\), we find \(l \approx 1 \text{ m}\).

The expression represents two perpendicular SHMs of the same frequency with a phase difference of \(\frac{\pi}{2}\). The resultant amplitude is \(A = \sqrt{\alpha^2 + \beta^2}\).

The maximum acceleration in SHM is given by \(a_{\text{max}} = \omega^2 A = (2\pi f)^2 A\). Substituting \(f = 4\text{ Hz}\) and \(A = 0.15\text{ m}\), we find \(a_{\text{max}} = (8\pi)^2 \times 0.15 \approx 94.65 \text{ m s}^{-2}\).

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 \(T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{K}}\).