If the period of oscillation of mass $m$ suspended from a spring is $2 \, s$,then the period of mass $4m$ will be .... $s$.

$A$ 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:

$A$ spring is stretched by $0.2 \ m$ when a mass of $0.5 \ kg$ is suspended to it. The time period of the spring when the $0.5 \ kg$ mass is replaced with a mass of $0.25 \ kg$ is (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $s$)

$A$ spring has a certain mass suspended from it and its period of vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $T_2$. The ratio of $T_2 / T_1$ is

$A$ block $P$ of mass $m$ is placed on a frictionless horizontal surface. Another block $Q$ of the same mass $m$ is kept on $P$ and connected to a wall with the help of a spring of spring constant $k$ as shown in the figure. $\mu_s$ is the coefficient of static friction between $P$ and $Q$. The blocks move together performing simple harmonic motion $(SHM)$ of amplitude $A$. The maximum value of the friction force between $P$ and $Q$ is

$A$ mass $m$ performs oscillations of period $T$ when suspended by a spring of force constant $K$. If the spring is cut into two equal parts and arranged in parallel,and the same mass $m$ is oscillated by them,then the new time period will be: