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Time period of a simple pendulum is $T$. The time taken to complete $5 / 8$ oscillations starting from mean position is $\frac{\alpha}{\beta} T$. The value of $\alpha$ 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
Let $T_1$ and $T_2$ be the time periods of two springs $A$ and $B$ when a mass $m$ is suspended from them separately. Now both the springs are connected in parallel and same mass $m$ is suspended with them. Now let $T$ be the time period in this position. Then
Two pendulum have time periods $T$ and $5T/4$. They start $SHM$ at the same time from the mean position. After how many oscillations of the smaller pendulum they will be again in the same phase
A spring has a certain mass suspended from it and its period for vertical oscillation is $T$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is now
A plank with a small block on top of it is under going vertical $SHM$ . Its period is $2\ sec$ . The minimum amplitude at which the block will separate from plank is
In damped oscillations, damping force is directly proportional to speed of oscillator. If amplitude becomes half of its maximum value in $1 \,s$, then after $2 \,s$ amplitude will be $\left(A_0-\right.$ initial amplitude)
The maximum potential energy of a block executing simple harmonic motion is $25\,J$. A is amplitude of oscillation. At $A / 2$, the kinetic energy of the block is $...............$
If the displacement of a particle executing $SHM $ is given by $y = 0.30\sin (220t + 0.64)$ in metre, then the frequency and maximum velocity of the particle is