The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is $10\,s^{-1}$. At, $t = 0$ the displacement is $5\, m$. What is the maximum acceleration ? The initial phase is $\frac{\pi }{4}$
JEE MAIN 2017, Diffcult
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 of a particle executing $S.H.M.$ is given by $x=0.01 \sin 100 \pi(t+0.05)$. The time period is ........ $s$View Solution
- 2The phase difference between two $SHM\,\,$ ${y_1}\, = \,10\,\sin \,\left( {10\pi t\, + \,\frac{\pi }{3}} \right)$ and ${y_2}\, = \,12\,\sin \,\left( {8\pi t\, + \,\frac{\pi }{4}} \right)$ at $t = 0.5\,s$ itView Solution
- 3A particle executing $SHM$ of amplitude $4\,cm$ and $T = 4s.$ The time taken by it to move from $+2\,cm$ to $+2\sqrt 3\,cm$ isView Solution
- 4If the length of simple pendulum is increased by $300\%$, then the time period will be increased by ..... $\%$View Solution
- 5View SolutionThe effective spring constant of two spring system as shown in figure will be
- 6A particle of mass $m$ moves in the potential energy $U$ shown above. The period of the motion when the particle has total energy $E$ isView Solution
- 7View SolutionThe frequency at which its kinetic energy change into potential energy is
- 8$Assertion :$ In simple harmonic motion, the velocity is maximum when the acceleration is minimum.View Solution
$Reason :$ Displacement and velocity of $S.H.M.$ differ in phase by $\frac{\pi }{2}$ - 9A mass $m$ is attached to two springs of same force constant $K$, as shown in following four arrangements. If $T_1, T_2, T_3$ and $T_4$ respectively be the time periods of oscillation in the following arrangements, in which case time period is maximum?View Solution
- 10The mass $M$ shown in the figure oscillates in simple harmonic motion with amplitude $A$. The amplitude of the point $P$ isView Solution
