The frequency of oscillation of a mass $m$ suspended by a spring is $v$. If the mass is reduced to one-fourth,what will be the new frequency of oscillation?

$A$ body of mass $64 \ g$ is made to oscillate turn by turn on two different springs $A$ and $B$. Spring $A$ and $B$ have force constants $4 \ N/m$ and $16 \ N/m$ respectively. If $T_{1}$ and $T_{2}$ are the periods of oscillation of springs $A$ and $B$ respectively,then $\frac{T_{1}+T_{2}}{T_{1}-T_{2}}$ will be:

$A$ spring has a certain mass suspended from it and its period for vertical oscillation is $T$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is now

Write the expression for the restoring force produced in a spring when a body attached to its end is pulled down by a small displacement $x$.

$A$ 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 $......$

$A$ spring of unstretched length $l$ has a mass $m$ with one end fixed to a rigid support. Assuming the spring to be made of a uniform wire,the kinetic energy possessed by it if its free end is pulled with uniform velocity $v$ is