Time period of a second’s pendulum is \(2\text{ s}\), the approximate length of its string is equal to (\(g = 10\text{ m/s}^2\))
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 \(l \approx 1 \text{ m}\).
A 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 \(3a\).
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
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
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
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
Comparing with the standard equation \(\frac{d^2y}{dt^2} + \omega^2 y = 0\), we find \(\omega = \sqrt{K}\). Thus, the time period is \(T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{K}}\).
A simple pendulum hanging freely stayed at rest in vertical, because in this position
A stable equilibrium state corresponds to a local minimum of the system's potential energy. For a simple pendulum, the lowest point is the vertical position, where potential energy is minimum.