Block $A$ is hanging from a vertical spring and it is at rest. Block $'B'$ strikes the block $'A'$ with velocity $v$ and stick to it. Then the velocity $v$ for which the spring just attains natural length is:
$g\sqrt[]{\frac{2m}{k}}$
$g\sqrt{\frac{6m}{k}}$
$g\sqrt{\frac{10m}{k}}$
$g\sqrt{\frac{14m}{k}}$
Two masses $m_1$ and $m_2$ are supended together by a massless spring of constant $k$. When the masses are in equilibrium, $m_1$ is removed without disturbing the system; the amplitude of vibration is
In the given figure, a mass $M$ is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is $k$. The mass oscillates on a frictionless surface with time period $T$ and amplitude $A$. When the mass is in equilibrium position, as shown in the figure, another mass $m$ is gently fixed upon it. The new amplitude of oscillation will be
A mass m oscillates with simple harmonic motion with frequency $f = \frac{\omega }{{2\pi }}$ and amplitude A on a spring with constant $K$ , therefore
How the period of oscillation depend on the mass of block attached to the end of spring ?
A spring executes $SHM$ with mass of $10\,kg$ attached to it. The force constant of spring is $10\,N/m$.If at any instant its velocity is $40\,cm/sec$, the displacement will be .... $m$ (where amplitude is $0.5\,m$)