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A uniform thin ring of radius $R$ and mass $m$ suspended in a vertical plane from a point in its circumference. Its time period of oscillation is ........
The kinetic energy of a particle executing $S.H.M.$ is $16\, J$ when it is in its mean position. If the amplitude of oscillations is $25\, cm$ and the mass of the particle is $5.12\, kg$, the time period of its oscillation is
The amplitude of a damped oscillator decreases to $0.9$ times its original magnitude in $5\ s$. In another $10\ s$ it will decrease to $\alpha $ times its original magnitude, where $\alpha $ equals
A simple pendulum has time period $T_1$. The point of suspension is now moved upward according to equation $y = k{t^2}$ where $k = 1\,m/se{c^2}$. If new time period is $T_2$ then ratio $\frac{{T_1^2}}{{T_2^2}}$ will be
silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of $10^{12} /sec$. What is the force constant of the bonds connecting one atom with the other? ................ $\mathrm{N/m}$ (Mole wt. of silver $= 108 $ andAvagadro number $= 6.02 \times 10^{23}$ $gm \ mole^{ -1}$ )
The amplitude and the periodic time of a $S.H.M.$ are $ 5\,cm$ and $6\,sec$ respectively. At a distance of $2.5\,cm$ away from the mean position, the phase will be
Two simple harmonic motions are represented by equations ${y_1} = 4\,\sin \,\left( {10t + \phi } \right)$ and ${y_2} = 5\,\cos \,10\,t$ What is the phase difference between their velocities?