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

A particle at the end of a spring executes simple harmonic motion with a period ${t_1}$, while the corresponding period for another spring is ${t_2}$. If the period of oscillation with the two springs in series is $T$, then

Two springs, of force constants $k_1$ and $k_2$ are connected to a mass $m$ as shown. The frequency of oscillation of the mass is $f$ If both $k_1$ and $k_2$ are made four times their original values, the frequency of oscillation becomes

Maximum amplitude(in $cm$) of $SHM$ so block A will not slip on block $B , K =100 N / m$

$A$ block of mass $M_1$ is hanged by a light spring of force constant $k$ to the top bar of a reverse Uframe of mass $M_2$ on the floor. The block is pooled down from its equilibrium position by $a$ distance $x$ and then released. Find the minimum value of $x$ such that the reverse $U$ -frame will leave the floor momentarily.

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 ........... $kg$