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 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 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, \(\omega^2 A = g \implies \left(\frac{2\pi}{T}\right)^2 A = g \implies 4\pi^2 A = 10 \implies 40 A = 10 \implies A = 0.25\text{ m}\).
If \( x = 2\sin\left(\frac{\pi}{2}t\right) \) represents the motion of a particle executing SHM, the maximum speed of the particle in \( \text{m s}^{-1} \) is (All parameters are in SI units)
Comparing the given equation with the standard SHM equation \( x = A\sin(\omega t) \), we get \( A = 2\text{ m} \) and \( \omega = \frac{\pi}{2}\text{ rad/s} \). The maximum speed is \( v_{\max} = A\omega = 2 \times \frac{\pi}{2} = \pi\text{ m/s} \).
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.
For the damped oscillator, if time taken for its amplitude of vibrations to drop to half of its initial value is \(T\) then time taken for amplitude to drop to one eighth amplitude is
Amplitude decays exponentially as \(A = A_0 e^{-\gamma t}\). Since it halves in time \(T\), to drop to \(1/8 = (1/2)^3\) of its initial value, it takes exactly \(3T\).