$A$ spring of length $l$ and force constant $k$ is attached to a mass $m$ and oscillates with frequency $f_1$. If the spring is cut into two equal parts and one part is attached to the same mass $m$ to oscillate,its frequency becomes $f_2$. Find the relation between $f_1$ and $f_2$.

The vertical extension in a light spring by a weight of $1\, kg$ suspended from it is $9.8\, cm$. What is the period of oscillation?

$A$ mass $0.9 \, kg$,attached to a horizontal spring,executes $SHM$ with an amplitude $A_{1}$. When this mass passes through its mean position,a smaller mass of $124 \, g$ is placed over it and both masses move together with amplitude $A_{2}$. If the ratio $\frac{A_{1}}{A_{2}}$ is $\frac{\alpha}{\alpha-1}$,then the value of $\alpha$ will be $......$

$A$ mass $m$ is suspended separately by two springs of spring constants $k_1$ and $k_2$. The time periods of oscillations in the two cases are $T_1$ and $T_2$ respectively. If the same mass $m$ is suspended by connecting the two springs in parallel (as shown in the figure),then the time period of the oscillation is $T$. The correct relation is:

$A$ force of $6.4 \, N$ stretches a vertical spring by $0.1 \, m$. The mass that must be suspended from the spring so that it oscillates with a period of $\left( \frac{\pi}{4} \right) \, s$ is ... $kg$.

When a body of mass $8 \,kg$ is attached to a spring balance, the reading of the balance is $20 \,cm$. Instead of $8 \,kg$, if another body of mass $M$ is suspended from the spring balance and is made to oscillate vertically, the time period of oscillation is $\frac{\pi}{5} \,s$, then the value of $M$ is (Acceleration due to gravity $= 10 \,m/s^2$) (in $\,kg$)