If the maximum velocity and maximum acceleration of a particle executing $SHM$ are equal in magnitude, the time period will be .... $\sec$
Easy
Download our appand get started for free
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.*
Similar Questions
- 1The displacement time graph of a particle executing $S.H.M.$ is as shown in the figureThe corresponding force-time graph of the particle isView Solution
- 2The amplitude and the periodic time of a $S.H.M.$ are $ 5\,cm$ and $6\,sec$ respectively. At a distance of $2.5\,cm$ away from the mean position, the phase will beView Solution
- 3The potential energy of a particle executing S.H.M. is $ 2.5\, J$, when its displacement is half of amplitude. The total energy of the particle be .... $J$View Solution
- 4A simple pendulum performs simple harmonic motion about $X = 0$ with an amplitude $A$ and time period $T$. The speed of the pendulum at $X = \frac{A}{2}$ will beView Solution
- 5The displacement of a particle from its mean position (in metre) is given by $y = 0.2\sin (10\pi t + 1.5\pi )\cos (10\pi t + 1.5\pi )$. The motion of particle isView Solution
- 6A simple pendulum is executing simple harmonic motion with a time period $T$. If the length of the pendulum is increased by $21\%$, the percentage increase in the time period of the pendulum of increased length is ..... $\%$View Solution
- 7The mass of a particle is $1\,\,kg$ and it is moving along $x-$ axis. The period of its small oscillation is $\frac {\pi }{2}$ . Its potential energy may beView Solution
- 8A particle executes $S.H.M.$ according to equation $x=10( cm ) \cos \left[2 \pi t+\frac{\pi}{2}\right]$, where $t$ is in second. The magnitude of the velocity of the particle at $t=\frac{1}{6} \,s$ will be .............. $cm / s$View Solution
- 9A particle is executing simple harmonic motion with a period of $T$ seconds and amplitude a metre. The shortest time it takes to reach a point $\frac{a}{{\sqrt 2 }}\,m$ from its mean position in seconds isView Solution
- 10A particle of mass $m$ is moving along a trajectory given byView Solution
$x = x_0 + a\, cos\,\omega_1 t$
$y = y_0 + b\, sin\,\omega_2t$
The torque, acing on the particle about the origin, at $t = 0$ is