The effective spring constant of the two-spring system as shown in the figure will be:

As shown in the figure, a block of weight $20 \,N$ is connected to the top of a smooth inclined plane by a massless spring of constant $8 \pi^2 \,Nm^{-1}$. If the block is pulled slightly from its mean position and released, the period of oscillations is (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,s$)

$A$ 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 $\pi = \frac{22}{7}$)

How does the period of oscillation depend on the mass of the block attached to the end of a spring?

On a smooth inclined plane,a body of mass $M$ is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has a force constant $K$,the period of oscillation of the body (assuming the springs are massless) is

$A$ body of mass $1 \,kg$ is attached to the lower end of a vertically suspended spring of force constant $600 \,N \,m^{-1}$. If another body of mass $0.5 \,kg$ moving vertically upward hits the suspended body with a velocity $3 \,m \,s^{-1}$ and gets embedded in it, then the frequency of the oscillation is