Step-by-Step Derivation:

1. Set up the Coordinate System and Define Motion

   • Let's place the origin (0, 0) at the point of contact between the two discs at time t=0.
   • The center of the left disc (Disc 1) is at C₁ = (-R, 0).
   • The center of the right disc (Disc 2) is at C₂ = (R, 0).
   • The left disc rotates counter-clockwise with angular velocity $\vec{\omega}_1 = \omega \hat{k}$.
   • The right disc rotates clockwise with angular velocity $\vec{\omega}_2 = -\omega \hat{k}$.
   • Point P is on Disc 1 and Point Q is on Disc 2. At t=0, both P and Q are at the origin (0, 0).

2. Velocity of Point P

   • The velocity of any point on a rotating body is given by $\vec{v} = \vec{v}_{center} + \vec{\omega} \times \vec{r}'$, where $\vec{r}'$ is the position vector from the center to the point.
   • The centers of the discs are stationary, so $\vec{v}_{C₁} = 0$.
   • At t=0, the position vector of P relative to C₁ is $\vec{r}'_{P,0} = (R, 0)$.
   • At time t, this vector rotates by an angle $\theta_1 = \omega t$ counter-clockwise.
   • The position vector of P relative to C₁ at time t is $\vec{r}'_P(t) = (R \cos(\omega t), R \sin(\omega t))$.
   • The velocity of P is $\vec{v}_P = \vec{\omega}_1 \times \vec{r}'_P(t)$. $\vec{v}_P = (\omega \hat{k}) \times (R \cos(\omega t) \hat{i} + R \sin(\omega t) \hat{j})$ $\vec{v}_P = R\omega \cos(\omega t) (\hat{k} \times \hat{i}) + R\omega \sin(\omega t) (\hat{k} \times \hat{j})$ $\vec{v}_P = R\omega \cos(\omega t) \hat{j} - R\omega \sin(\omega t) \hat{i} = (-R\omega \sin(\omega t), R\omega \cos(\omega t))$

3. Velocity of Point Q

   • The center C₂ is also stationary, so $\vec{v}_{C₂} = 0$.
   • At t=0, the position vector of Q relative to C₂ is $\vec{r}'_{Q,0} = (-R, 0)$.
   • At time t, this vector rotates by an angle $\theta_2 = -\omega t$ (clockwise).
   • The position vector of Q relative to C₂ at time t is $\vec{r}'_Q(t) = (-R \cos(-\omega t), -R \sin(-\omega t)) = (-R \cos(\omega t), R \sin(\omega t))$.
   • The velocity of Q is $\vec{v}_Q = \vec{\omega}_2 \times \vec{r}'_Q(t)$. $\vec{v}_Q = (-\omega \hat{k}) \times (-R \cos(\omega t) \hat{i} + R \sin(\omega t) \hat{j})$ $\vec{v}_Q = R\omega \cos(\omega t) (\hat{k} \times \hat{i}) - R\omega \sin(\omega t) (-\hat{k} \times \hat{j})$ $\vec{v}_Q = R\omega \cos(\omega t) \hat{j} + R\omega \sin(\omega t) \hat{i} = (R\omega \sin(\omega t), R\omega \cos(\omega t))$

4. Relative Velocity and Speed

   • The relative velocity of P with respect to Q is $\vec{v}_{PQ} = \vec{v}_P - \vec{v}_Q$. $\vec{v}_{PQ} = (-R\omega \sin(\omega t) - R\omega \sin(\omega t)) \hat{i} + (R\omega \cos(\omega t) - R\omega \cos(\omega t)) \hat{j}$ $\vec{v}_{PQ} = -2R\omega \sin(\omega t) \hat{i}$
   • The relative speed $v_r$ is the magnitude of the relative velocity vector $\vec{v}_{PQ}$. $v_r = |\vec{v}_{PQ}| = |-2R\omega \sin(\omega t)| = 2R\omega |\sin(\omega t)|$

5. Analyze the Function $v_r(t)$

   • The time period of rotation of the discs is $T = 2\pi / \omega$.
   • We need to analyze the graph of $v_r(t) = 2R\omega |\sin(\omega t)|$ for t from 0 to T.
   • Let's check the value of $v_r$ at key points in the interval [0, T]:
     • At t = 0: $v_r = 2R\omega |\sin(0)| = 0$.
     • At t = T/4: $\omega t = (2\pi/T) \times (T/4) = \pi/2$. $v_r = 2R\omega |\sin(\pi/2)| = 2R\omega$ (Maximum value).
     • At t = T/2: $\omega t = (2\pi/T) \times (T/2) = \pi$. $v_r = 2R\omega |\sin(\pi)| = 0$.
     • At t = 3T/4: $\omega t = (2\pi/T) \times (3T/4) = 3\pi/2$. $v_r = 2R\omega |\sin(3\pi/2)| = 2R\omega |-1| = 2R\omega$ (Maximum value).
     • At t = T: $\omega t = (2\pi/T) \times T = 2\pi$. $v_r = 2R\omega |\sin(2\pi)| = 0$.

6. Conclusion

   • The graph of $v_r(t)$ starts at 0, increases sinusoidally to a maximum of $2R\omega$ at t=T/4, decreases sinusoidally back to 0 at t=T/2.
   • It then repeats this pattern for the second half of the period, from t=T/2 to t=T.
   • This behavior corresponds to a rectified sine wave. The period of $v_r(t)$ is T/2, so there are two full cycles (humps) within one rotational period T.
   • This description perfectly matches the graph shown in option A.