Two objects A and B of equal mass are suspended from two springs constants `k_(A)` and `k_(B)` if the objects oscillate vertically in such a manner that their maximum kinetic energies are equal, then the ratio of their amplitudes is

Options

A. `(K_(B))/(K_(A))`
B. `sqrt((K_(B))/(K_(A)))`
C. `(K_(A))/(K_(B))`
D. `sqrt((K_(A))/(K_(B)))`

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the relationship between kinetic energy and amplitude
The maximum kinetic energy (KE) of an object in simple harmonic motion is given by the formula:
\[ KE_{max} = \frac{1}{2} k A^2 \]
where \( k \) is the spring constant and \( A \) is the amplitude of oscillation.

### Step 2: Write down the equations for both objects
For object A, the maximum kinetic energy is:
\[ KE_{A} = \frac{1}{2} k_A A_A^2 \]
For object B, the maximum kinetic energy is:
\[ KE_{B} = \frac{1}{2} k_B A_B^2 \]

### Step 3: Set the maximum kinetic energies equal
According to the problem, the maximum kinetic energies of both objects are equal:
\[ \frac{1}{2} k_A A_A^2 = \frac{1}{2} k_B A_B^2 \]

### Step 4: Simplify the equation
We can cancel the \( \frac{1}{2} \) from both sides:
\[ k_A A_A^2 = k_B A_B^2 \]

### Step 5: Rearrange to find the ratio of amplitudes
Rearranging the equation gives:
\[ \frac{A_A^2}{A_B^2} = \frac{k_B}{k_A} \]

### Step 6: Take the square root to find the ratio of amplitudes
Taking the square root of both sides results in:
\[ \frac{A_A}{A_B} = \sqrt{\frac{k_B}{k_A}} \]

### Final Result
Thus, the ratio of the amplitudes of objects A and B is:
\[ \frac{A_A}{A_B} = \sqrt{\frac{k_B}{k_A}} \]