Two springs of force constant $K$ and $2K$ are connected to a mass as shown below. The frequency of oscillation of the mass is
$\frac{{3K}}{M}$
$\pi \,\sqrt {\frac{{6M}}{K}} $
$\frac{1}{{2\pi }}\,\sqrt {\frac{{3K}}{M}} $
$\frac{1}{\pi }\,\sqrt {\frac{K}{{6M}}} $
The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If length of spring is cut to one third then the same mass oscillates with frequency $v_2$, then
A bar of mass $m$ is suspended horizontally on two vertical springs of spring constant $k$ and $3k$ . The bar bounces up and down while remaining horizontal. Find the time period of oscillation of the bar (Neglect mass of springs and friction everywhere).
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
The springs shown are identical. When $A = 4kg$, the elongation of spring is $1\, cm$. If $B = 6\,kg$, the elongation produced by it is ..... $ cm$
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