$A$ particle of mass $0.50 \ kg$ executes simple harmonic motion under force $F = -50 \ (N/m) x$. The time period of oscillation is $\frac{x}{35} \ s$. The value of $x$ is . . . . . (Given $\pi = \frac{22}{7}$)

When a body of mass $8 \,kg$ is attached to a spring balance, the reading of the balance is $20 \,cm$. Instead of $8 \,kg$, if another body of mass $M$ is suspended from the spring balance and is made to oscillate vertically, the time period of oscillation is $\frac{\pi}{5} \,s$, then the value of $M$ is (Acceleration due to gravity $= 10 \,m/s^2$) (in $\,kg$)

$A$ mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude $A$. Its speed as it passes through the equilibrium position is $V$. If extended $2A$ and released,the speed of the mass passing through the equilibrium position will be

$A$ particle of mass $m$ is attached to three identical springs $A, B$ and $C$ each of force constant $k$ as shown in the figure. If the particle of mass $m$ is pushed slightly against the spring $A$ and released,then the time period of oscillations is

The potential energy of a particle of mass $0.02 \ kg$ moving along the $x$-axis is given by $V = A x(x-4) \ J$, where $x$ is in metres and $A$ is a constant. Which of the following is/are correct statement$(s)$?

The frequency of oscillation of a mass $m$ suspended by a spring is $v$. If the mass is reduced to one-fourth,what will be the new frequency of oscillation?