Two ends of a string AB are attached to two rigid walls. The string is oscillating transeversely in its first overtone. When the string is passing through its equilibrium position kinetic energy of an element of the string. 

All particles of the string, except the one at C, are oscillating with same frequency. Amplitude of oscillation ‘A’ at D is maximum. Now maximum kinetic energy ${\mathrm{K}}_{\max }=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{A}}^2$. Where ‘m’ is the mass of an element. $\text{'}\mathrm{\omega }\text{'}$ and ‘A’ are zero for the particle at C. Hence option-(1) is correct.