A $1 \,kg$ block attached to a spring vibrates with a frequency of $1\, Hz$ on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an $8\, kg$ block placed on the same table. So, the frequency of vibration of the $8\, kg$ block is ..... $Hz$

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 $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.

When a mass $m$ is attached to a spring, it normally extends by $0.2\, m$. The mass $m$ is given a slight addition extension and released, then its time period will be

What will be the force constant of the spring system shown in the figure

Two particles of mass $m$ are constrained to move along two horizontal frictionless rails that make an angle $2\theta $ with respect to each other. They are connected by a spring with spring constant $k$ . The angular frequency of small oscillations for the motion where the two masses always stay parallel to each other (that is the distance between the meeting point of the rails and each particle is equal) is 

A uniform spring of force constant $k$ is cut into two pieces, the lengths of which are in the ratio $1 : 2$. The ratio of the force constants of the shorter and the longer pieces is