A mass $m$ is suspended separately by two different springs of spring constant $K_1$ and $K_2$ gives the time-period ${t_1}$ and ${t_2}$ respectively. If same mass $m$ is connected by both springs as shown in figure then time-period $t$ is given by the relation
$t = {t_1} + {t_2}$
$t = \frac{{{t_1}.{t_2}}}{{{t_1} + {t_2}}}$
${t^2} = {t_1}^2 + {t_2}^2$
${t^{ - 2}} = {t_1}^{ - 2} + {t_2}^{ - 2}$
Two bodies of masses $1\, kg$ and $4\, kg$ are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency $25\, rad/s$, and amplitude $1.6\, cm$ while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is ..... $N$ ( take $g = 10\, ms^{-2}$)
One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance ${x_0}$ from the black. The spring is then compressed by $2{x_0}$ and released. The time taken to strike the wall is
Two springs with spring constants ${K_1} = 1500\,N/m$ and ${K_2} = 3000\,N/m$ are stretched by the same force. The ratio of potential energy stored in spring will be
A block of mass $m$ attached to massless spring is performing oscillatory motion of amplitude $'A'$ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become $fA.$ The value of $f$ is
For the damped oscillator shown in Figure the mass mof the block is $200\; g , k=90 \;N m ^{-1}$ and the damping constant $b$ is $40 \;g s ^{-1} .$ Calculate
$(a)$ the period of oscillation,
$(b)$ time taken for its amplitude of vibrations to drop to half of Its inittal value, and
$(c)$ the time taken for its mechanical energy to drop to half its initial value.