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A man weighing $60\ kg$ stands on the horizontal platform of a spring balance. The platform starts executing simple harmonic motion of amplitude $0.1\ m$ and frequency $\frac{2}{\pi } Hz$. Which of the following staements is correct
The motion of a mass on a spring, with spring constant ${K}$ is as shown in figure. The equation of motion is given by $x(t)= A sin \omega t+ Bcos\omega t$ with $\omega=\sqrt{\frac{K}{m}}$ Suppose that at time $t=0$, the position of mass is $x(0)$ and velocity $v(0)$, then its displacement can also be represented as $x(t)=C \cos (\omega t-\phi)$, where $C$ and $\phi$ are
$Assertion :$ In simple harmonic motion, the motion is to and fro and periodic
$Reason :$ Velocity of the particle $(v) = \omega \sqrt {k^2 - x^2}$ (where $x$ is the displacement).
${T}_{0}$ is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to $\frac{1}{16}$ times of its initial value, the modified time
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.
The time period of a particle executing $S.H.M.$ is $8 \,s$. At $t=0$ it is at the mean position. The ratio of distance covered by the particle in $1^{\text {st }}$ second to the $2^{\text {nd }}$ second is .............. $s$