A particle of mass m is executing osciallations about the origin on the x-axis with amplitude A. its potential energy is given as `U(x)=alphax^(4)`, where `alpha` is a positive constant. The x-coordinate of mass where potential energy is one-third the kinetic energy of particle is

Options

A. `+-A/2`
B. `+-A/sqrt(2)`
C. `+-A/3`
D. `+-A/sqrt(3)`

Answer

To solve the problem step by step, we will follow these logical steps:

### Step 1: Understand the relationship between potential energy (U) and kinetic energy (K).
Given that the potential energy \( U(x) \) is one-third of the kinetic energy \( K \), we can express this relationship mathematically:
\[ U = \frac{1}{3} K \]
This implies that:
\[ K = 3U \]

### Step 2: Write the expression for total mechanical energy (E).
The total mechanical energy \( E \) of the system is conserved and can be expressed as:
\[ E = U + K \]
Substituting \( K \) from the previous step:
\[ E = U + 3U = 4U \]

### Step 3: Determine the total mechanical energy at maximum amplitude.
At maximum amplitude \( x = A \), the potential energy is:
\[ U(A) = \alpha A^4 \]
At this point, the kinetic energy is zero (since the particle is momentarily at rest), so the total mechanical energy is:
\[ E = U(A) = \alpha A^4 \]

### Step 4: Set the total mechanical energy equal to the expression derived from potential and kinetic energy.
From Step 2, we have:
\[ E = 4U \]
Setting this equal to the expression for total mechanical energy at maximum amplitude:
\[ \alpha A^4 = 4U \]

### Step 5: Substitute the expression for potential energy at position \( x \).
At position \( x \), the potential energy is:
\[ U(x) = \alpha x^4 \]
Substituting this into the equation from Step 4 gives:
\[ \alpha A^4 = 4(\alpha x^4) \]

### Step 6: Simplify the equation.
Canceling \( \alpha \) (since it is a positive constant) from both sides:
\[ A^4 = 4x^4 \]

### Step 7: Solve for \( x \).
Rearranging gives:
\[ x^4 = \frac{A^4}{4} \]
Taking the fourth root of both sides:
\[ x = \pm \frac{A}{\sqrt{2}} \]

### Conclusion:
The x-coordinate where the potential energy is one-third the kinetic energy is:
\[ x = \pm \frac{A}{\sqrt{2}} \]