$A$ spring is stretched by $0.20\, m$ when a mass of $0.50\, kg$ is suspended. When a mass of $0.25\, kg$ is suspended,its period of oscillation will be .... $sec$ $(g = 10\, m/s^2)$

$A$ particle of mass $1 \, \text{kg}$ is hanging from a spring of force constant $100 \, \text{N/m}$. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period $T$. The time when the kinetic energy and potential energy of the system will become equal is $\frac{T}{x}$. The value of $x$ is ..... .

$A$ 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?

Find the maximum amplitude (in $cm$) of the $SHM$ such that block $A$ will not slip on block $B$. Given: spring constant $K = 100 \ N/m$,mass of block $A$ $m_A = 0.25 \ kg$,mass of block $B$ $m_B = 1.25 \ kg$,and coefficient of friction between $A$ and $B$ is $\mu = 0.4$. Take $g = 10 \ m/s^2$.

$A$ wire of length $L$,cross-sectional area $A$,and Young's modulus $Y$ is suspended,and a spring of force constant $k$ is attached to its lower end. If a mass $m$ is suspended from the spring and set into oscillations,what will be the time period of the system?

$A$ mass $m$ is suspended separately by two springs of spring constants $k_1$ and $k_2$. The time periods of oscillations in the two cases are $T_1$ and $T_2$ respectively. If the same mass $m$ is suspended by connecting the two springs in parallel (as shown in the figure),then the time period of the oscillation is $T$. The correct relation is: