Oscillations

Hint: Both springs can be considered equivalent to a parallel combination of springs.

Step: Find the time period of the combination of the springs.
As both springs are connected in parallel then the effective spring constant is given by; 
\({k}_{eq}{=}{k}_{1}{+}{k}_{2}\)
The time period of the spring combination is given by;  
\({T}{=}{2}\mathit{\pi}\sqrt{\frac{m}{k_{eq}}}\)
\(\Rightarrow {T}{=}{2}\mathit{\pi}\sqrt{\frac{m}{{k}_{1}{+}{k}_{2}}}\)
Hence, option (3) is the correct answer.
 

 

Hint:  \(KE_{max} =PE_{max}\)

Step:

Apply the conservation of mechanical energy.
The kinetic energy for a particle is given by;
\(K = K_0 \sin^2(\omega t)\) 
\(K_i+U_i = K_f+U_f\) [When \(K_i = 0, \Rightarrow U_i = U_{\text{max}}\) Similarly \(U_f =0 \Rightarrow K_f = K_{\text{max}}\)]
Therefore maximum value of total energy is given as;
\(TE=KE _{\max }=K_0\)
As the total energy of the simple harmonic motion is constant therefore we can say that the maximum value of potential energy will be;
 \(\Rightarrow U_{max} =K_{max}\)
\(\Rightarrow U_{max} =K_0\)
Hence, option (1) is the correct answer.