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A particle is excuting a simple harmonic motion. Its maximum acceleration is $\alpha $ and maximum velocity is $\beta $. Then its frequency of vibration will be
This is the position time graph of a mass on spring. What can you say about the velocity and force at the instant indicated by dashed line ? (positive direction is to the right)
A particle is executing simple harmonic motion with a time period $T.$ At time $t = 0$, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like
A mass on a vertical spring begins its motion at rest at $y = 0\ cm$. It reaches a maximum height of $y = 10\ cm$. The two forces acting on the mass are gravity and the spring force. The graph of its kinetic energy ($KE$) versus position is given below. Net force on the mass varies with $y$ as
A $1.00 \times {10^{ - 20}}kg$ particle is vibrating with simple harmonic motion with a period of $1.00 \times {10^{ - 5}}sec$ and a maximum speed of $1.00 \times {10^3}m/s$. The maximum displacement of the particle is
In figure $(A),$ mass ' $2 m$ ' is fixed on mass ' $m$ ' which is attached to two springs of spring constant $k$. In figure $(B),$ mass ' $m$ ' is attached to two spring of spring constant ' $k$ ' and ' $2 k$ '. If mass ' $m$ ' in $(A)$ and $(B)$ are displaced by distance ' $x$ ' horizontally and then released, then time period $T_{1}$ and $T_{2}$ corresponding to $(A)$ and $(B)$ respectively follow the relation.
Vertical displacement of a plank with a body of mass $'m'$ on it is varying according to law $y = \sin \omega t + \cos \omega t.$ The minimum value of $\omega $ for which the mass just breaks off the plank and the moment it occurs first after $t = 0$ are given by : ( $y$ is positive vertically upwards)