Assertion (A): Vibration of polyatomic molecules is not simple harmonic motion.
Reason (R): The vibrations are superposition of SHMs of different frequency.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
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.
Equation of SHM of a particle whose amplitude is 0.1 m and frequency is 25 Hz with an initial phase of \(\frac{\pi}{4}\) radians is
1. \(x = 0.1 sin \left(2\pi t + \frac{\pi}{4}\right)\)
2. \(x = 0.1 sin \left(2\pi t - \frac{\pi}{4}\right)\)
3. \(x = 0.1 sin \left(50\pi t + \frac{\pi}{4}\right)\)
4. \(x = 0.1 sin (50\pi t)\)
View Answer
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 \(x = 0.1 sin \left(50\pi t + \frac{\pi}{4}\right)\).
The displacement of a harmonic oscillator is given by \(x = \alpha \sin\omega t + \beta \cos\omega t\). The amplitude of the oscillation is
1. \(\alpha + \beta\)
2. \(\alpha^2 + \beta^2\)
3. \(\sqrt{\alpha^2 + \beta^2}\)
4. \(\sqrt{\alpha^2 + \beta^2 + 2\alpha\beta}\)
View Answer
The expression represents two perpendicular SHMs of the same frequency with a phase difference of \(\frac{\pi}{2}\). The resultant amplitude is \(A = \sqrt{\alpha^2 + \beta^2}\).
A body is vibrating with SHM of amplitude 15 cm and frequency 4 Hz. The maximum value of acceleration is
1. \(10.65 \text{ m s}^{-2}\)
2. \(80.52 \text{ m s}^{-2}\)
3. \(94.65 \text{ m s}^{-2}\)
4. \(68.52 \text{ m s}^{-2}\)
View Answer
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 \(a_{\text{max}} = (8\pi)^2 \times 0.15 \approx 94.65 \text{ m s}^{-2}\).
The differential equation of motion of a particle executing SHM is \(\frac{d^2y}{dt^2} + Ky = 0\) where K is positive constant. The time period of the oscillation is given by
1. \(\frac{2\pi}{K}\)
2. \(2\pi K\)
3. \(\frac{2\pi}{\sqrt{K}}\)
4. \(2\pi\sqrt{K}\)
View Answer
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 displacement of a harmonic oscillator is given by \(x = \alpha \sin\omega t + \beta \cos\omega t\). The amplitude of the oscillation is
1. \(\alpha + \beta\)
2. \(\alpha^2 + \beta^2\)
3. \(\sqrt{\alpha^2 + \beta^2}\)
4. \(\sqrt{\alpha^2 + \beta^2 + 2\alpha\beta}\)
View Answer
Since the two perpendicular components have a phase difference of \(\frac{\pi}{2}\), the net amplitude is \(A = \sqrt{\alpha^2 + \beta^2}\).
A body is vibrating with SHM of amplitude \(15\text{ cm}\) and frequency \(4\text{ Hz}\). The maximum value of acceleration is
1. \(10.65\text{ m s}^{-2}\)
2. \(80.52\text{ m s}^{-2}\)
3. \(94.65\text{ m s}^{-2}\)
4. \(68.52\text{ m s}^{-2}\)
View Answer
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 \(a_{\text{max}} = 4\pi^2 (16)(0.15) \approx 94.65\text{ m/s}^2\).
The differential equation of motion of a particle executing SHM is \(\frac{d^2y}{dt^2} + Ky = 0\) where \(K\) is positive constant. The time period of the oscillation is given by
1. \(\frac{2\pi}{K}\)
2. \(2\pi K\)
3. \(\frac{2\pi}{\sqrt{K}}\)
4. \(2\pi\sqrt{K}\)
View Answer
Comparing with the standard equation \(\frac{d^2y}{dt^2} + \omega^2 y = 0\), we find \(\omega = \sqrt{K}\). Thus, the time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{K}}\).