When a mass $m$ is attached to a spring it oscillates with period $4 \,s$. When an additional mass of $2 \,kg$ is attached to a spring, time period increases by $1 \,s$. The value of $m$ is ........... $kg$

A weightless spring of length $60\, cm$ and force constant $200\, N/m$ is kept straight and unstretched on a smooth horizontal table and its ends are rigidly fixed. A mass of $0.25\, kg$ is attached at the middle of the spring and is slightly displaced along the length. The time period of the oscillation of the mass is

A spring whose unstretched length is $\ell $ has a force constant $k$. The spring is cut into two pieces of unstretched lengths $\ell_1$ and $\ell_2$ where, $\ell_1 = n\ell_2$ and $n$ is an integer. The ratio $k_1/k_2$ of the corresponding force constants, $k_1$ and $k_2$ will be

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.

To make the frequency double of a spring oscillator, we have to

A mass $M$, attached to a horizontal spring, executes S.H.M. with amplitude $A_1$. When the mass $M$ passes through its mean position then a smaller mass $m$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\frac{{{A_1}}}{{{A_2}}}$ is