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
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A point mass oscillates along the x-axis according to the law $x=x_0cos$$\left( {\omega t - \frac{\pi }{4}} \right)$ If the acceleration of the particle is written as $a=Acos$$\left( {\omega t + \delta } \right)$ then
A particle of mass $m$ undergoes oscillations about $x=0$ in a potential given by $V(x)-\frac{1}{2} k x^2-V_0 \cos \left(\frac{x}{a}\right)$, where $V_0, k, a$ are constants. If the amplitude of oscillation is much smaller than $a$, the time period is given by
A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ${\omega _0}$ - An external force $F (t)$ proportional to $\cos \omega \,t((\omega \ne {\omega _0})$ is applied to the oscillator. The time displacement of the oscillator will be proportional to
The angular velocity and the amplitude of a simple pendulum is $\omega $ and $a$ respectively. At a displacement $X$ from the mean position if its kinetic energy is $T$ and potential energy is $V$, then the ratio of $T$ to $V$ is