Choose the correct statement.

Options

A. If `vecA + vecB= vecA - vecB` then `vecB` is a null vector
B. If `|vecA + vecB| = |vecA-vecB|` , then `vecA` and `vecB` are perpendicular vectors
C. both of the above
D. none of the above

Answer

To solve the problem of finding the time period for small oscillations of a ring connected to a spring, we can follow these steps:

### Step 1: Understand the System
We have a ring of mass \( m \) and radius \( a \) connected to a spring with spring constant \( k \). The ring can rotate about a hinge in a vertical plane. When the ring is in a horizontal position, it is in equilibrium.

**Hint:** Visualize the setup and identify the forces acting on the ring when it is displaced.

### Step 2: Displacement and Arc Length
When the ring is displaced by a small angle \( \theta \), the horizontal displacement \( x \) of the ring can be approximated as:
\[ x = 2a \theta \]
This is because the arc length for a small angle in radians is given by \( \text{arc length} = r \theta \).

**Hint:** Remember that for small angles, \( \sin(\theta) \approx \theta \).

### Step 3: Extension in the Spring
The extension \( x \) in the spring due to the displacement is equal to the horizontal displacement of the ring. The force exerted by the spring is given by Hooke's Law:
\[ F_{\text{spring}} = kx = k(2a\theta) \]

**Hint:** Identify how the spring force acts to restore the ring to its equilibrium position.

### Step 4: Calculate the Torque
The torque \( \tau \) due to the spring force about the hinge is:
\[ \tau = F_{\text{spring}} \cdot \text{perpendicular distance} = k(2a\theta) \cdot (2a) = 4ak\theta \]

**Hint:** Recall that torque is the product of force and the perpendicular distance from the pivot point.

### Step 5: Moment of Inertia
Using the perpendicular axis theorem, the moment of inertia \( I \) of the ring about the hinge is:
\[ I = \frac{1}{2} m (2a)^2 = 2ma^2 \]

**Hint:** Ensure you apply the correct formula for the moment of inertia for a ring.

### Step 6: Equation of Motion
For small oscillations, the angular displacement \( \theta \) leads to a restoring torque proportional to \( \theta \):
\[ \tau = -I \frac{d^2\theta}{dt^2} \]
Substituting the expressions for torque and moment of inertia:
\[ 4ak\theta = -2ma^2 \frac{d^2\theta}{dt^2} \]

**Hint:** This is a second-order differential equation resembling simple harmonic motion.

### Step 7: Time Period of Oscillation
Rearranging the equation gives:
\[ \frac{d^2\theta}{dt^2} + \frac{4ak}{2ma^2}\theta = 0 \]
This indicates simple harmonic motion with angular frequency:
\[ \omega^2 = \frac{4ak}{2ma^2} \]
Thus, the time period \( T \) is given by:
\[ T = 2\pi \sqrt{\frac{I}{\tau}} = 2\pi \sqrt{\frac{2ma^2}{4ak}} = 2\pi \sqrt{\frac{ma^2}{2ak}} = 2\pi \sqrt{\frac{m}{2k}} \]

**Hint:** The time period \( T \) is derived from the relationship between angular frequency and time period in simple harmonic motion.

### Final Answer
The time period for small oscillations of the ring is:
\[ T = 2\pi \sqrt{\frac{m}{2k}} \]