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Two simple pendulums of length $1\, m$ and $4\, m$ respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed number of oscillations equal to
A particle of mass m is under the influence of a force $F$ which varies with the displacement $x$ according to the relation $F = - kx + {F_0}$ in which $k$ and ${F_0}$ are constants. The particle when disturbed will oscillate
The period of a simple pendulum measured inside a stationary lift is found to be $T$. If the lift starts accelerating upwards with acceleration of $g/3,$ then the time period of the pendulum is
Time period of a particle executing $SHM$ is $8\, sec.$ At $t = 0$ it is at the mean position. The ratio of the distance covered by the particle in the $1^{st}$ second to the $2^{nd}$ second is :
A pendulum is suspended in a lift and its period of oscillation when the lift is stationary is $T_0$. What must be the acceleration of the lift for the period of oscillation of the pendulum to be $T_0/2$ ?
A cuboidal piece of wood has dimensions $a, b$ and $c$. Its relative density is $d$. It is floating in a large body of water such that side a is vertical. It is pushed down a bit and released. The time period of $SHM$ executed by it is :
A particle performing $SHM$ is found at its equilibrium at $t = 1\,sec$. and it is found to have a speed of $0.25\,m/s$ at $t = 2\,sec$. If the period of oscillation is $6\,sec$. Calculate amplitude of oscillation