A spring having a spring constant $‘K’$ is loaded with a mass $‘m’$. The spring is cut into two equal parts and one of these is loaded again with the same mass. The new spring constant is

A 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?

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

Initially system is in equilibrium. Time period of $SHM$ of block in vertical direction is

In figure $(A),$ mass ' $2 m$ ' is fixed on mass ' $m$ ' which is attached to two springs of spring constant $k$. In figure $(B),$ mass ' $m$ ' is attached to two spring of spring constant ' $k$ ' and ' $2 k$ '. If mass ' $m$ ' in $(A)$ and $(B)$ are displaced by distance ' $x$ ' horizontally and then released, then time period $T_{1}$ and $T_{2}$ corresponding to $(A)$ and $(B)$ respectively follow the relation.

The mass $M$ shown in the figure oscillates in simple harmonic motion with amplitude $A$. The amplitude of the point $P$ is