Thank you for reporting, we will resolve it shortly
Q.
Two springs are connected to a block of mass $M$ placed on a frictionless surface as shown below. If both the springs have a spring constant $k$, the frequency of oscillation of the block is
Let when the oscillating mass is at a distance $x$ towards right from its equilibrium position, the instantaneous extensions in the springs of force constants $k$, is
$x=x_{1}+x_{2}$
Since, the springs are in series the restoring force exerted by each spring on mass $m$ is same. Then
$ F =-k x_{1}=-k x_{2} $
$ \therefore x_{1} =-\frac{F}{k}, x_{2}=-\frac{F}{k} $
and $x=x_{1}+x_{2}=-F\left(\frac{1}{k}+\frac{1}{k}\right)=\frac{-2 F}{k}$
$\therefore F=-\frac{k}{2} x$
$\Rightarrow $ Effective force constant is $\frac{k}{2}$.
Hence, frequency of oscillation is
$n=\frac{1}{2 \pi} \sqrt{\frac{k}{2 M}}$
