Two masses ${m_1}$ and ${m_2}$ are suspended together by a massless spring of constant k. When the masses are in equilibrium, ${m_1}$ is removed without disturbing the system. Then the angular frequency of oscillation of ${m_2}$ is
$\sqrt {\frac{k}{{{m_1}}}} $
$\sqrt {\frac{k}{{{m_2}}}} $
$\sqrt {\frac{k}{{{m_1} + {m_2}}}} $
$\sqrt {\frac{k}{{{m_1}{m_2}}}} $
A body executes simple harmonic motion under the action of a force $F_1$ with a time period $(4/5)\, sec$. If the force is changed to $F_2$ it executes $SHM$ with time period $(3/5)\, sec$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body, its time period (in $seconds$ ) is
A mass $M$ is suspended from a light spring. An additional mass m added displaces the spring further by a distance $x$. Now the combined mass will oscillate on the spring with period
The springs in figure. $A$ and $B$ are identical but length in $A$ is three times that in $B$. The ratio of period $T_A/T_B$ is
A mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of time period $T$. If the mass is increased by m, the time period becomes $5T/3$. Then the ratio of $m/M$ is
In arrangement given in figure, if the block of mass m is displaced, the frequency is given by