Time period of a simple pendulum is $T$. The angular displacement for amplitude is $\beta$. How much time the bob of pendulum will take to move from equilibrium position $O$ to $A$, making an angle $\alpha$ at the support
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated in the figure. The simple harmonic motion of the $x-$ projection of the radius vector of the rotating particle $P$ is
The bob of a pendulum was released from a horizontal position. The length of the pendulum is $10 \mathrm{~m}$. If it dissipates $10 \%$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is : [Use, $\mathrm{g}: 10 \mathrm{~ms}^{-2}$ ]
Abody performs simple harmonic oscillations along the straight line $ABCDE$ with $C$ as the midpoint of $AE.$ Its kinetic energies at $B$ and $D$ are each one fourth of its maximum value. If $AE = 2R,$ the distance between $B$ and $D$ is
A body executing simple harmonic motion has a maximum acceleration equal to $ 24\,metres/se{c^2} $ and maximum velocity equal to $ 16\;metres/sec $. The amplitude of the simple harmonic motion is
Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to $A$ and $T,$ respectively. At time $t=0$ one particle has displacement $A$ while the other one has displacement $\frac {-A}{2}$ and they are moving towards each other. If they cross each other at time $t,$ then $t$ is
The total energy of a particle executing $S.H.M.$ is $80 \,J$. What is the potential energy when the particle is at a distance of $\frac{3}{4}$ of amplitude from the mean position..... $J$
The potential energy of a particle of mass $100 \,g$ moving along $x$-axis is given by $U=5 x(x-4)$, where $x$ is in metre. The period of oscillation is .................
A particle executes $SHM$ on a straight line path. The amplitude of oscillation is $2\, cm.$ When the displacement of the particle from the mean position is $1\, cm,$ the numerical value of magnitude of acceleration is equal to the numerical value of magnitude of velocity. The frequency of $SHM$ (in $second^{-1}$) is :