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
$m_1g / k$
$m_2g / k$
$\frac{{\left( {{m_1} + {m_2}} \right)\,g}}{k}$
$\frac{{\left( {{m_2} - {m_1}} \right)\,g}}{k}$
A spring has a certain mass suspended from it and its period for vertical oscillation is $T$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is now
A body of mass $0.01 kg$ executes simple harmonic motion $(S.H.M.)$ about $x = 0$ under the influence of a force shown below : The period of the $S.H.M.$ is .... $s$
One-forth length of a spring of force constant $K$ is cut away. The force constant of the remaining spring will be
What is condition for a body suspended at the end of a spring having simple harmonic oscillation ?
A $1\,kg$ mass is attached to a spring of force constant $600\,N / m$ and rests on a smooth horizontal surface with other end of the spring tied to wall as shown in figure. A second mass of $0.5\,kg$ slides along the surface towards the first at $3\,m / s$. If the masses make a perfectly inelastic collision, then find amplitude and time period of oscillation of combined mass.