Two springs,of force constants $k_1$ and $k_2$ are connected to a mass $m$ as shown. The frequency of oscillation of the mass is $f$. If both $k_1$ and $k_2$ are made four times their original values,the frequency of oscillation becomes

$A$ particle of mass $200 \,g$ executes $S.H.M.$ The restoring force is provided by a spring of force constant $80 \,N/m$. The time period of oscillations is .... $s$.

The time period of oscillation for the given combination will be:

The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to half,the same mass oscillates with frequency $v_2$. The value of $v_2/v_1$ is . . . . . . .

$A$ spring-mass system (mass $m$,spring constant $k$,and natural length $\ell_{0}$) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the center of the disc. If the disc,together with the spring-mass system,rotates about its axis with an angular velocity $\omega$ (where $k >> m \omega^{2}$),the relative change in the length of the spring is best given by which option?

The potential energy of a particle of mass $4 \, kg$ in motion along the $x$-axis is given by $U = 4(1 - \cos 4x) \, J$. The time period of the particle for small oscillations $(\sin \theta \simeq \theta)$ is $\left(\frac{\pi}{K}\right) \, s$. The value of $K$ is .......