$A$ simple pendulum with a bob of mass $m$ and length $x$ is held in position at an angle $\theta_1$ and then at an angle $\theta_2$ with the vertical. When released from these positions,the speeds with which it passes the lowest position are $v_1$ and $v_2$ respectively. Then,the ratio $\frac{v_1}{v_2}$ is .............

• A
  $\frac{1-\cos \theta_1}{1-\cos \theta_2}$
• B
  $\sqrt{\frac{1-\cos \theta_1}{1-\cos \theta_2}}$
• C
  $\sqrt{\frac{2 g x(1-\cos \theta_1)}{1-\cos \theta_2}}$
• D
  $\sqrt{\frac{1-\cos \theta_1}{2 g x(1-\cos \theta_2)}}$