The vertical extension in a light spring by a weight of $1\, kg$ suspended from the wire is $9.8\, cm$. The period of oscillation

The motion of a mass on a spring, with spring constant ${K}$ is as shown in figure. The equation of motion is given by $x(t)= A sin \omega t+ Bcos\omega t$ with $\omega=\sqrt{\frac{K}{m}}$ Suppose that at time $t=0$, the position of mass is $x(0)$ and velocity $v(0)$, then its displacement can also be represented as $x(t)=C \cos (\omega t-\phi)$, where $C$ and $\phi$ are

The springs shown are identical. When $A = 4kg$, the elongation of spring is $1\, cm$. If $B = 6\,kg$, the elongation produced by it is  ….. $ cm$

A block is placed on a frictionless horizontal table. The mass of the block is m and springs are attached on either side with force constants ${K_1}$ and ${K_2}$. If the block is displaced a little and left to oscillate, then the angular frequency of oscillation will be

A particle of mass $m$ is attached to one end of a mass-less spring of force constant $k$, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u_0$. When the speed of the particle is $0.5 u_0$, it collies elastically with a rigid wall. After this collision :

$(A)$ the speed of the particle when it returns to its equilibrium position is $u_0$.

$(B)$ the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{ m }{ k }}$.

$(C)$ the time at which the maximum compression of the spring occurs is $t =\frac{4 \pi}{3} \sqrt{\frac{ m }{ k }}$.

$(D)$ the time at which the particle passes througout the equilibrium position for the second time is $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.