A particle is placed at the lowest point of a smooth wire frame in the shape of a parabola, lying in the vertical $xy-$ plane having equation $x^2 = 5y$ $(x, y$ are in meter). After slight displacement, the particle is set free. Find angular frequency of oscillation.....$rad/s$ (in $rad/sec$ ) (take $g = 10\ m/s^2$ )
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Two pendulums have time periods $T$ and $\frac{{5T}}{4}.$They start $S.H.M.$ at the same time from the mean position. What will be the phase difference between them after the bigger pendulum has complete one oscillation ..... $^o$
Two bodies performing $S.H.M.$ have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is
A $1.00 \times 10^{-20} \,kg$ particle is vibrating under simple harmonic motion with a period of $1.00 \times 10^{-5} \,s$ and with a maximum speed of $1.00 \times 10^3 \,m / s$. The maximum displacement of particle from mean position is .......... $mm$
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
A point mass oscillates along the x-axis according to the law $x=x_0cos$$\left( {\omega t - \frac{\pi }{4}} \right)$ If the acceleration of the particle is written as $a=Acos$$\left( {\omega t + \delta } \right)$ then
The angular velocity and the amplitude of a simple pendulum is $\omega $ and $a$ respectively. At a displacement $X$ from the mean position if its kinetic energy is $T$ and potential energy is $V$, then the ratio of $T$ to $V$ is
In simple harmonic motion, the total mechanical energy of given system is E. If mass of oscillating particle $P$ is doubled then the new energy of the system for same amplitude is:
The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of $\pi $ results in the displacement of the particle along