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
$\frac{{Mg}}{{{K_1} + {K_2}}}$
$\frac{{Mg\,({K_1} + {K_2})}}{{{K_1}{K_2}}}$
$\frac{{Mg\,{K_1}{K_2}}}{{{K_1} + {K_2}}}$
$\frac{{{K_1} + {K_2}}}{{{K_1}{K_2}Mg}}$
Find the ratio of time periods of two identical springs if they are first joined in series $\&$ then in parallel $\&$ a mass $m$ is suspended from them :
A spring of force constant $k$ is cut into lengths of ratio $1:2:3$ . They are connected in series and the new force constant is $k'$ . Then they are connected in parallel and force constant is $k''$ . Then $k':k''$ is
What is the period of small oscillations of the block of mass $m$ if the springs are ideal and pulleys are massless ?
The vertical extension in a light spring by a weight of $1\, kg$ suspended from the wire is $9.8\, cm$. The period of oscillation
For the damped oscillator shown in Figure the mass mof the block is $200\; g , k=90 \;N m ^{-1}$ and the damping constant $b$ is $40 \;g s ^{-1} .$ Calculate
$(a)$ the period of oscillation,
$(b)$ time taken for its amplitude of vibrations to drop to half of Its inittal value, and
$(c)$ the time taken for its mechanical energy to drop to half its initial value.