Find the maximum amplitude (in $cm$) of the $SHM$ such that block $A$ will not slip on block $B$. Given: spring constant $K = 100 \ N/m$,mass of block $A$ $m_A = 0.25 \ kg$,mass of block $B$ $m_B = 1.25 \ kg$,and coefficient of friction between $A$ and $B$ is $\mu = 0.4$. Take $g = 10 \ m/s^2$.

$A$ block of mass $m$ attached to a 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 becomes $fA$. The value of $f$ is

The force-deformation equation for a nonlinear spring fixed at one end is $F = 4x^{1/2}$,where $F$ is the force (expressed in newtons) applied at the other end and $x$ is the deformation expressed in meters.

$A$ solid cylinder of mass $m$ and volume $V$ is suspended from the ceiling by a spring of spring constant $k$. It has a cross-sectional area $A$. It is submerged in a liquid of density $\rho$ up to half its length. If a small block of mass $M_0$ is kept at the center of the top, the amplitude of small oscillation will be:

When a mass $m$ is connected individually with two springs $S_1$ and $S_2$,the oscillation frequencies are $n_1$ and $n_2$. If the mass $m$ is attached to the springs as shown in the figure,the oscillation frequency would be

$A$ mass attached to a spring is free to oscillate,with angular velocity $\omega$,in a horizontal plane without friction or damping. It is pulled to a distance $x_{0}$ and pushed towards the centre with a velocity $v_{0}$ at time $t=0$. Determine the amplitude of the resulting oscillations in terms of the parameters $\omega, x_{0}$ and $v_{0}$. [Hint: Start with the equation $x=A \cos (\omega t+\theta)$ and note that the initial velocity is negative.]