When a mass \(m\) is connected individually to two springs \(S_1\) and \(S_2,\) the oscillation frequencies are \(\nu_1\) and \(\nu_2.\) If the same mass is attached to the two springs as shown in the figure, the oscillation frequency would be: 

         
1. \(v_2+v_2\)

2. \(\sqrt{v_1^2+v_2^2}\)

3. \(\frac{1}{v_1}+\frac{1}{v_1}^{-1}\)

4. \(\sqrt{v_1^2-v_2^2}\)

2. Hint: Two springs are in parallel.
     
Step 1: Find the equivalent frequency of the system.
Consider the diagram,
k=Equivaent spring constant=k1+k2
The time period of oscillation of the spring block-system,
                         T=2πmkeq=2πmk1+k2
                        v=1T=12π×k1+k2m                       ...(i)
Step 2: Find the individual frequencies.
When the mass is connected to the springs individually:
                                  
                             v1=12πk1m                     ...(ii)
v2=12πk2m                     ...(iii)
Using equations (i), (ii) and (iii)....
v=12π×4π2ν12+4π2ν22=ν12+ν22