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 will analyze the motion of the toy cart under the influence of Hooke's law and centripetal force.

### 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 \( T \). The elongation of the string when the cart is at radius \( 2a \) is:
\[ x_1 = 2a - a = a \]

### Step 2: Apply centripetal force
The centripetal force required for the cart to move in a circle is provided by the spring force, which follows Hooke's law:
\[ F = kx_1 = k \cdot a \]
This force must equal the centripetal force, which is given by:
\[ F_c = \frac{mv^2}{r} = \frac{mv^2}{2a} \]
Setting these equal gives:
\[ k \cdot a = \frac{mv^2}{2a} \]

### Step 3: Relate speed and time period
The speed \( v \) can be expressed in terms of the time period \( T \):
\[ v = \frac{2\pi(2a)}{T} = \frac{4\pi a}{T} \]
Substituting this into the centripetal force equation:
\[ k \cdot a = \frac{m\left(\frac{4\pi a}{T}\right)^2}{2a} \]
This simplifies to:
\[ k \cdot a = \frac{m \cdot 16\pi^2 a}{2T^2} \]
\[ k = \frac{8m\pi^2}{T^2} \]

### Step 4: Analyze the new conditions
Now, the toy cart is moved to a radius of \( 3a \) with the same time period \( T \). The new elongation \( x_2 \) is:
\[ x_2 = 3a - a = 2a \]

### Step 5: Set up the new centripetal force equation
The new centripetal force must also be equal to the spring force:
\[ F = kx_2 = k \cdot 2a \]
And the centripetal force is:
\[ F_c = \frac{mv'^2}{r} = \frac{mv'^2}{3a} \]
Setting these equal gives:
\[ k \cdot 2a = \frac{mv'^2}{3a} \]

### Step 6: Substitute \( k \) from earlier
Substituting \( k \) from the previous step:
\[ \frac{8m\pi^2}{T^2} \cdot 2a = \frac{mv'^2}{3a} \]
This simplifies to:
\[ \frac{16m\pi^2 a}{T^2} = \frac{mv'^2}{3a} \]
Cancelling \( m \) and rearranging gives:
\[ v'^2 = \frac{48\pi^2 a^2}{T^2} \]

### Step 7: Relate \( v' \) to the new time period \( T' \)
The new speed \( v' \) can also be expressed in terms of the new time period \( T' \):
\[ v' = \frac{2\pi(3a)}{T'} \]
Squaring this gives:
\[ v'^2 = \frac{4\pi^2(3a)^2}{T'^2} = \frac{36\pi^2 a^2}{T'^2} \]

### Step 8: Set the two expressions for \( v'^2 \) equal
Equating the two expressions for \( v'^2 \):
\[ \frac{48\pi^2 a^2}{T^2} = \frac{36\pi^2 a^2}{T'^2} \]
Cancelling \( \pi^2 a^2 \) gives:
\[ \frac{48}{T^2} = \frac{36}{T'^2} \]
Cross-multiplying gives:
\[ 48T'^2 = 36T^2 \]
Thus,
\[ T'^2 = \frac{36}{48}T^2 = \frac{3}{4}T^2 \]
Taking the square root gives:
\[ T' = \frac{\sqrt{3}}{2}T \]

### Final Result
The new time period \( T' \) is:
\[ T' = \frac{\sqrt{3}}{2}T \]