What is the period of small oscillations of the block of mass $m$ if the springs are ideal and pulleys are massless ?
$\frac{\pi }{2}\sqrt {\frac{m}{k}}$
$\frac{\pi }{2}\sqrt {\frac{m}{2k}}$
$\frac{\pi }{2}\sqrt {\frac{2m}{k}}$
$\pi \sqrt {\frac{m}{{2k}}}$
A block of mass $m$ hangs from three springs having same spring constant $k$. If the mass is slightly displaced downwards, the time period of oscillation will be
Force constant of a spring is $K$ . If half part is detached then force constant of the remaining spring will be
A particle of mass $m$ is attached to one end of a mass-less spring of force constant $k$, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u_0$. When the speed of the particle is $0.5 u_0$, it collies elastically with a rigid wall. After this collision :
$(A)$ the speed of the particle when it returns to its equilibrium position is $u_0$.
$(B)$ the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{ m }{ k }}$.
$(C)$ the time at which the maximum compression of the spring occurs is $t =\frac{4 \pi}{3} \sqrt{\frac{ m }{ k }}$.
$(D)$ the time at which the particle passes througout the equilibrium position for the second time is $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.
A spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of
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 $k_2$.