$A$ mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of period $T$. If the mass is increased by $m$,the time period becomes $\frac{5T}{3}$. What is the ratio $\left(\frac{M}{m}\right)$?

One end of a long metallic wire of length $L$ is tied to the ceiling. The other end is tied to a massless spring of spring constant $K$. $A$ mass $m$ hangs freely from the free end of the spring. The area of cross-section and Young's modulus of the wire are $A$ and $Y$ respectively. If the mass is slightly pulled down and released,it will oscillate with a time period $T$ equal to

$A$ mass $M$ is suspended by two springs of force constants $K_1$ and $K_2$ respectively,as shown in the diagram. The total elongation (stretch) of the two springs is

$A$ mass $m = 1.0\,kg$ is placed on a flat pan attached to a vertical spring fixed on the ground. The mass of the spring and the pan is negligible. When pressed slightly and released,the mass executes simple harmonic motion. The spring constant is $k = 500\,N/m$. What is the amplitude $A$ of the motion,so that the mass $m$ tends to get detached from the pan? (Take $g = 10\,m/s^2$).

$A$ particle is suspended from a vertical spring which is executing $S.H.M.$ of frequency $5 \ Hz$. The spring is unstretched at the highest point of oscillation. What is the maximum speed of the particle? (Take $g = 10 \ m/s^2$)

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?