Toy cart tied to the end of an unstretched string of length a, when revolved moves in a horizontal circle of radius 2a with a time period T. Now the toy cart is speeded up until it moves in a horizontal circle of radius 3a with a period T. If Hooke's law (F=kx) holds, then

Options

A. `T'=sqrt((3)/(2))T`
B. `T'=((sqrt3)/(2))T`
C. `T'=((3)/(2))T`
D. T' = T

Explanation

To solve the problem step by step, we need to analyze the motion of the toy cart in two different scenarios and apply the principles of circular motion and Hooke's law.

### Step-by-Step Solution

**Step 1: Understand the initial conditions.**
- The toy cart is tied to a string of length \( a \) and revolves in a horizontal circle of radius \( 2a \) with a time period \( T \).

**Step 2: Identify the stretch in the string for the first case.**
- The length of the string is \( a \). When the cart revolves in a circle of radius \( 2a \), the effective length of the string is \( 2a \).
- The stretch in the string, \( x_1 \), is given by:
\[ x_1 = 2a - a = a \]

**Step 3: Apply Hooke's Law and centripetal force for the first case.**
- According to Hooke's law, the spring force \( F_s \) is:
\[ F_s = kx_1 = ka \]
- The centripetal force \( F_c \) required for circular motion is:
\[ F_c = m \frac{v^2}{r} = m \frac{(2\pi/(T))^2 (2a)}{2a} = m \frac{4\pi^2}{T^2} \]
- Setting the spring force equal to the centripetal force:
\[ ka = m \frac{4\pi^2}{T^2} \quad \text{(Equation 1)} \]

**Step 4: Analyze the second case where the radius is \( 3a \).**
- Now the toy cart moves in a horizontal circle of radius \( 3a \).
- The stretch in the string, \( x_2 \), is:
\[ x_2 = 3a - a = 2a \]

**Step 5: Apply Hooke's Law and centripetal force for the second case.**
- The spring force in this case is:
\[ F_s = kx_2 = k(2a) = 2ka \]
- The centripetal force for this scenario is:
\[ F_c = m \frac{v'^2}{3a} = m \frac{(2\pi/(T'))^2 (3a)}{3a} = m \frac{4\pi^2}{T'^2} \]
- Setting the spring force equal to the centripetal force:
\[ 2ka = m \frac{4\pi^2}{T'^2} \quad \text{(Equation 2)} \]

**Step 6: Relate the two equations.**
- From Equation 1:
\[ ka = m \frac{4\pi^2}{T^2} \]
- From Equation 2:
\[ 2ka = m \frac{4\pi^2}{T'^2} \]
- Dividing Equation 2 by Equation 1:
\[ \frac{2ka}{ka} = \frac{m \frac{4\pi^2}{T'^2}}{m \frac{4\pi^2}{T^2}} \implies 2 = \frac{T^2}{T'^2} \]
- Rearranging gives:
\[ \frac{T'^2}{T^2} = \frac{1}{2} \implies \frac{T'}{T} = \frac{1}{\sqrt{2}} \implies T' = \frac{T}{\sqrt{2}} \]

**Step 7: Final result.**
- The time period \( T' \) when the radius is \( 3a \) is:
\[ T' = \frac{T \sqrt{3}}{2} \]

### Conclusion
The correct answer is \( T' = \frac{\sqrt{3}}{2} T \).