$A$ 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 ... $kg$.

$A$ mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period $T$. If the mass is increased by $m$,then the time period becomes $\frac{5}{4}T$. The ratio of $\frac{m}{M}$ is

The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to half,the same mass oscillates with frequency $v_2$. The value of $v_2/v_1$ is . . . . . . .

$A$ spring is stretched by $0.40 \ m$ when a mass of $0.6 \ kg$ is suspended from it. The period of oscillations of the spring loaded by $255 \ g$ and put to oscillations is close to $(g = 10 \ m \ s^{-2})$. (in $s$)

The angular frequency of a block of mass $0.1 \ kg$ oscillating with the help of a spring of force constant $2.5 \ Nm^{-1}$ is

The time period of oscillation for the given combination will be: