Two masses $M_{A}$ and $M_{B}$ are hung from two strings of length $l_{A}$ and $l_{B}$ respectively. They are executing SHM with frequency relation $f_{A}=2 f_{B}$, then relation
$l_{A}=4 l_{B},$ does not depend on mass
$l_{A}=\frac{l_{B}}{4},$ does not depend on mass
$l_A=2 l_B$ and $M_A=2M_B$
$l_{A}=\frac{l_{B}}{2}$ and $M_{A}=\frac{M_{B}}{2}$
Let $T_1$ and $T_2$ be the time periods of two springs $A$ and $B$ when a mass $m$ is suspended from them separately. Now both the springs are connected in parallel and same mass $m$ is suspended with them. Now let $T$ be the time period in this position. Then
A uniform cylinder of length $L$ and mass $M$ having cross-sectional area $A$ is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density $\sigma $ at equilibrium position. When the cylinder is given a downward push and released, it starts oscillating vertically with a small amplitude. The time period $T$ of the oscillations of the cylinder will be
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 :-
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.
A mass $m$ is suspended separately by two different springs of spring constant $K_1$ and $K_2$ gives the time-period ${t_1}$ and ${t_2}$ respectively. If same mass $m$ is connected by both springs as shown in figure then time-period $t$ is given by the relation