In This post you will get answer of Q.
If f is the frequency when mass m is attached to a spring of spring constant k, then new frequency for this arrangement, is
Solution:
The frequency of oscillation of spring mass system is given by
$ , , , , , , , , , , , f=frac{1}{2pi}sqrt{frac{k}{m}}$
For the given arrangement
$ , , , , , , , , , , , , , k’ = k_1+k_2=k+k=2k$
Hence, frequency of oscillation is
$ , , , , f’ =frac{1}{2pi} sqrt{frac{k’}{m}} Rightarrow , f’ =frac{1}{2pi} sqrt{frac{2k}{m}}=sqrt 2 f$
Solution:
The frequency of oscillation of spring mass system is given by
$ , , , , , , , , , , , f=frac{1}{2pi}sqrt{frac{k}{m}}$
For the given arrangement
$ , , , , , , , , , , , , , k’ = k_1+k_2=k+k=2k$
Hence, frequency of oscillation is
$ , , , , f’ =frac{1}{2pi} sqrt{frac{k’}{m}} Rightarrow , f’ =frac{1}{2pi} sqrt{frac{2k}{m}}=sqrt 2 f$