The frequency of oscillation of the springs shown in the figure will be
$\frac{1}{{2\pi }}\sqrt {\frac{K}{m}} $
$\frac{1}{{2\pi }}\sqrt {\frac{{({K_1} + {K_2})m}}{{{K_1}{K_2}}}} $
$2\pi \sqrt {\frac{K}{m}} $
$\frac{1}{{2\pi }}\sqrt {\frac{{{K_1}{K_2}}}{{m({K_1} + {K_2})}}} $
A spring with $10$ coils has spring constant $k$. It is exactly cut into two halves, then each of these new springs will have a spring constant
Two springs of force constant $K$ and $2K$ are connected to a mass as shown below. The frequency of oscillation of the mass is
A spring block system in horizontal oscillation has a time-period $T$. Now the spring is cut into four equal parts and the block is re-connected with one of the parts. The new time period of vertical oscillation will be
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 simple harmonic oscillations with a time period $T$. If the mass is increased by m then the time period becomes $\left( {\frac{5}{4}T} \right)$. The ratio of $\frac{m}{{M}}$ is
A $5\; kg$ collar is attached to a spring of spring constant $500\;N m ^{-1} .$ It slides without friction over a hortzontal rod. The collar is displaced from its equilibrium position by $10.0\; cm$ and released. Calculate
$(a)$ the period of oscillation.
$(b)$ the maximum speed and
$(c)$ maximum acceleration of the collar.