If $f$ is the frequency when mass m is attached to a spring of spring constant k and undergoes SHM, then new frequency for this arrangement is $\sqrt{x}f$. The value of  $x=$?

Given

Frequency of oscillation = f

Mass of body to be attached = m

Spring constant = k

We know that the frequency of oscillation of spring mass system is given by

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$      

Both springs are identical, new spring constant for two parallel springs,

$k^1=k_1+k_2$   $\Rightarrow$ $k+k=2k$            

Frequency of oscillation,

$f^1=\frac{1}{2\pi }\sqrt{\frac{k^1}{m}}$ $\Rightarrow$$f^1=\frac{1}{2\pi }\sqrt{\frac{2k}{m}}$ $\Rightarrow$$f^1=\sqrt{2}f$