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On a planet a freely falling body takes $2 \,sec$ when it is dropped from a height of $8 \,m$, the time period of simple pendulum of length $1\, m$ on that planet is ..... $\sec$
If $x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m}$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
Two simple pendulums of length $0.5\, m$ and $2.0\, m$ respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed .... oscillations.
A mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of time period $T$. If the mass is increased by m, the time period becomes $5T/3$. Then the ratio of $m/M$ is
A function is represented by equation$y = A\,\cos \,\omega t\,\cos \,2\omega t + A\,\sin \,\omega t\,\sin \,2\omega t$.
Than the nature of the function is
Two oscillating systems; a simple pendulum and a vertical spring-mass-system have same time period of motion on the surface of the Earth. If both are taken to the moon, then-