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
$\frac{{{k_1}}}{{{k_2}}}$
$\sqrt {\frac{{{k_1}}}{{{k_2}}}} $
$\frac{{{k_2}}}{{{k_1}}}$
$\sqrt {\frac{{{k_2}}}{{{k_1}}}} $
A particle of mass $m$ is attached to three identical springs $A, B$ and $C$ each of force constant $ k$ a shown in figure. If the particle of mass $m$ is pushed slightly against the spring $A$ and released then the time period of oscillations is
A body of mass $5\; kg$ hangs from a spring and oscillates with a time period of $2\pi $ seconds. If the ball is removed, the length of the spring will decrease by
A mass attached to a spring is free to oscillate, with angular velocity $\omega,$ in a hortzontal plane without friction or damping. It is pulled to a distance $x_{0}$ and pushed towards the centre with a velocity $v_{ o }$ at time $t=0 .$ Determine the amplitude of the resulting oscillations in terms of the parameters $\omega, x_{0}$ and $v_{ o } .$ [Hint: Start with the equation $x=a \cos (\omega t+\theta)$ and note that the initial velocity is negative.]
In the figure given below. a block of mass $M =490\,g$ placed on a frictionless table is connected with two springs having same spring constant $\left( K =2 N m ^{-1}\right)$. If the block is horizontally displaced through ' $X$ 'm then the number of complete oscillations it will make in $14 \pi$ seconds will be $.........$
How the period of oscillation depend on the mass of block attached to the end of spring ?