$A$ 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 time period $T$. If the mass is increased by $m$,the time period becomes $5T/3$. Then the ratio of $m/M$ is

$A$ block of mass $m$ has two similar rubber ribbons attached to it as shown in the figure. The force constant of each rubber ribbon is $K$ and the surface is frictionless. The block is displaced from the mean position by $x$ and released. At the mean position,the ribbons are undeformed. The vibration period is

On a frictionless horizontal plane,a bob of mass $m=0.1 \text{ kg}$ is attached to a spring with natural length $l_0=0.1 \text{ m}$. The spring constant is $k_1=0.009 \text{ N/m}$ when the length of the spring $l > l_0$ and is $k_2=0.016 \text{ N/m}$ when $l < l_0$. Initially,the bob is released from $l=0.15 \text{ m}$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T=(n \pi) \text{ s}$,then the integer closest to $n$ is:

$A$ mass $m$ attached to the free end of a spring executes $SHM$ with a period of $1\; s$. If the mass is increased by $3\; kg$,the period of oscillation increases by $1\; s$. The value of mass $m$ is $..............kg$.

$A$ spring-mass system vibrates such that the mass travels on a surface with a coefficient of friction $\mu$. The mass is released after compressing the spring by a distance $a$ and it travels up to a distance $b$ after its equilibrium position. Then,while traveling from $x = -a$ to $x = b$,the reduction in its amplitude will be:

$A$ block of mass $m$ is at rest on another block of the same mass as shown in the figure. The lower block is attached to a spring. What is the maximum amplitude of motion such that both blocks remain in contact?