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
$m_1g / k$
$m_2g / k$
$\frac{{\left( {{m_1} + {m_2}} \right)\,g}}{k}$
$\frac{{\left( {{m_2} - {m_1}} \right)\,g}}{k}$
Two masses $m_1=1 \,kg$ and $m_2=0.5 \,kg$ are suspended together by a massless spring of spring constant $12.5 \,Nm ^{-1}$. When masses are in equilibrium $m_1$ is removed without disturbing the system. New amplitude of oscillation will be .......... $cm$
A mass $m$ is attached to two springs as shown in figure. The spring constants of two springs are $K _1$ and $K _2$. For the frictionless surface, the time period of oscillation of mass $m$ is
A weightless spring which has a force constant oscillates with frequency $n$ when a mass $m$ is suspended from it. The spring is cut into two equal halves and a mass $2m $ is suspended from it. The frequency of oscillation will now become
The spring mass system oscillating horizontally. What will be the effect on the time period if the spring is made to oscillate vertically ?
What is the period of small oscillations of the block of mass $m$ if the springs are ideal and pulleys are massless ?