Question 51:
easyAssertion (A): Vibration of polyatomic molecules is not simple harmonic motion.
Reason (R): The vibrations are superposition of SHMs of different frequency.
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.
Question 52:
moderateEquation 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
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 \(x = 0.1 sin \left(50\pi t + \frac{\pi}{4}\right)\).
Question 53:
moderateThe displacement of a harmonic oscillator is given by \(x = \alpha \sin\omega t + \beta \cos\omega t\). The amplitude of the oscillation is
The expression represents two perpendicular SHMs of the same frequency with a phase difference of \(\frac{\pi}{2}\). The resultant amplitude is \(A = \sqrt{\alpha^2 + \beta^2}\).
Question 54:
moderateA body is vibrating with SHM of amplitude 15 cm and frequency 4 Hz. The maximum value of acceleration is
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 \(a_{\text{max}} = (8\pi)^2 \times 0.15 \approx 94.65 \text{ m s}^{-2}\).
Question 55:
easyThe 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
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 \(T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{K}}\).
Question 56:
easyA particle moves according to equation, \(x = a \cos \frac{\pi t}{2}\) . The distance covered by it in the time interval between t = 0 to t = 3 s is
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 \(3a\).
Question 57:
easyThe displacement of a harmonic oscillator is given by \(x = \alpha \sin\omega t + \beta \cos\omega t\). The amplitude of the oscillation is
Since the two perpendicular components have a phase difference of \(\frac{\pi}{2}\), the net amplitude is \(A = \sqrt{\alpha^2 + \beta^2}\).
Question 58:
easyA body is vibrating with SHM of amplitude \(15\text{ cm}\) and frequency \(4\text{ Hz}\). The maximum value of acceleration is
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 \(a_{\text{max}} = 4\pi^2 (16)(0.15) \approx 94.65\text{ m/s}^2\).
Question 59:
easyThe 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
Comparing with the standard equation \(\frac{d^2y}{dt^2} + \omega^2 y = 0\), we find \(\omega = \sqrt{K}\). Thus, the time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{K}}\).
Question 60:
moderateA block is resting on a piston which is moving vertically executing SHM of period 1 s. At what minimum amplitude of motion, will the block and piston separate? (take \(\pi^2 = 10\))
Separation occurs when the maximum downward acceleration of the piston equals \(g\). Thus, \(\omega^2 A = g \implies \left(\frac{2\pi}{T}\right)^2 A = g \implies 4\pi^2 A = 10 \implies 40 A = 10 \implies A = 0.25\text{ m}\).