A mass $M$ is suspended by two springs of force constants $K_1$ and $K_2$ respectively as shown in the diagram. The total elongation (stretch) of the two springs is
$\frac{{Mg}}{{{K_1} + {K_2}}}$
$\frac{{Mg\,({K_1} + {K_2})}}{{{K_1}{K_2}}}$
$\frac{{Mg\,{K_1}{K_2}}}{{{K_1} + {K_2}}}$
$\frac{{{K_1} + {K_2}}}{{{K_1}{K_2}Mg}}$
A force of $6.4\, N$ stretches a vertical spring by $0.1 \,m$. The mass that must be suspended from the spring so that it oscillates with a period of $\left( {\frac{\pi }{4}} \right)sec$. is ... $kg$
An assembly of identical spring-mass systems is placed on a smooth horizontal surface as shown. Initially the springs are relaxed. The left mass is displaced to the left while the right mass is displaced to the right and released. The resulting collision is elastic. The time period of the oscillations of the system is :-
Two bodies $M$ and $N $ of equal masses are suspended from two separate massless springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude $M$ to that of $N$ is
Two springs with spring constants ${K_1} = 1500\,N/m$ and ${K_2} = 3000\,N/m$ are stretched by the same force. The ratio of potential energy stored in spring will be
Two springs having spring constant $k_1$ and $k_2$ is connected in series, its resultant spring constant will be $2\,unit$. Now if they connected in parallel its resultant spring constant will be $9\,unit$, then find the value of $k_1$ and $k_2$.