The displacement of a damped harmonic oscillator is given by $x\left( t \right) = {e^{ - 0.1\,t}}\,\cos \left( {10\pi t + \varphi } \right)$ The time taken for its amplitude of vibration to drop to half of its initial value is close to .... $s$
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Two particles are in $SHM$ in a straight line. Amplitude $A$ and time period $T$ of both the particles are equal. At time $t=0,$ one particle is at displacement $y_1= +A$ and the other at $y_2= -A/2,$ and they are approaching towards each other. After what time they cross each other ?
A uniform cylinder of length $L$ and mass $M$ having cross-sectional area $A$ is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density $\sigma $ at equilibrium position. When the cylinder is given a downward push and released, it starts oscillating vertically with a small amplitude. The time period $T$ of the oscillations of the cylinder will be
A particle is executing $S.H.M.$ with time period $T^{\prime}$. If time period of its total mechanical energy is $T$ then $\frac{T^{\prime}}{T}$ is ........
The amplitude of a damped oscillator becomes one third in $2\, sec$. If its amplitude after $6\, sec$ is $1/n$ times the original amplitude then the value of $n$ is
A simple harmonic oscillator has a period of $0.01 \,sec$ and an amplitude of $0.2\, m$. The magnitude of the velocity in $m{\sec ^{ - 1}}$ at the centre of oscillation is
Displacement-time equation of a particle executing $SHM$ is $x\, = \,A\,\sin \,\left( {\omega t\, + \,\frac{\pi }{6}} \right)$ Time taken by the particle to go directly from $x\, = \, - \frac{A}{2}$ to $x\, = \, + \frac{A}{2}$ is