An assembly of identical spring-mass systems is placed on a smooth horizontal surface as shown. At this instant, the springs are relaxed. The left mass is displaced to the/left and theiright mass is displaced to the right by same distance and released. The resulting collision is elastic. The time period of the oscillations of system is
$2 \pi \sqrt{\frac{2 m}{k}}$
$2 \pi \sqrt{\frac{m}{2 k}}$
$\pi \sqrt{\frac{m}{k}}$
$2 \pi \sqrt{\frac{m}{k}}$
The period of oscillation of a mass $M$ suspended from a spring of negligible mass is $T$. If along with it another mass $M$ is also suspended , the period of oscillation will now be
A $2\, Kg$ block moving with $10\, m/s$ strikes a spring of constant $\pi ^2 N/m$ attached to $2\, Kg$ block at rest kept on a smooth floor. The time for which rear moving block remain in contact with spring will be ... $\sec$
$Assertion :$ The time-period of pendulum, on a satellite orbiting the earth is infinity.
$Reason :$ Time-period of a pendulum is inversely proportional to $\sqrt g$
A block with mass $M$ is connected by a massless spring with stiffiess constant $k$ to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude $A$ about an equilibrium position $x_0$. Consider two cases: ($i$) when the block is at $x_0$; and ($ii$) when the block is at $x=x_0+A$. In both the cases, a perticle with mass $m$ is placed on the mass $M$ ?
($A$) The amplitude of oscillation in the first case changes by a factor of $\sqrt{\frac{M}{m+M}}$, whereas in the second case it remains unchanged
($B$) The final time period of oscillation in both the cases is same
($C$) The total energy decreases in both the cases
($D$) The instantaneous speed at $x_0$ of the combined masses decreases in both the cases
Define simple pendulum and the length of pendulum.