There exist a potential field
U(x)= `U_(0)(1-cos alpha x)` .where
`U_(0) & alpha` are constants. A mass m is
located in this field. The time
period of oscillation of this mass is
found to be proportional to
`m^(a)U_(0)^(b)alpha^(c).` Now find the product of abc=?

Explanation

To solve the problem, we need to determine the time period of oscillation for a mass \( m \) located in the potential field given by:

\[ U(x) = U_0(1 - \cos(\alpha x)) \]

### Step 1: Calculate the Force
The force \( F \) acting on the mass can be derived from the potential energy using the relation:

\[ F = -\frac{dU}{dx} \]

Calculating the derivative:

\[ F = -\frac{d}{dx}[U_0(1 - \cos(\alpha x))] = -U_0 \cdot \frac{d}{dx}[-\cos(\alpha x)] = -U_0 \cdot \alpha \sin(\alpha x) \]

### Step 2: Approximate for Small Displacements
For small displacements, we can use the small angle approximation \( \sin(\theta) \approx \theta \):

\[ F \approx -U_0 \cdot \alpha \cdot (\alpha x) = -U_0 \alpha^2 x \]

This resembles Hooke's law, \( F = -kx \), where \( k = U_0 \alpha^2 \).

### Step 3: Determine the Time Period
The time period \( T \) of a simple harmonic oscillator is given by:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

Substituting \( k \):

\[ T = 2\pi \sqrt{\frac{m}{U_0 \alpha^2}} \]

### Step 4: Express \( T \) in Terms of \( m, U_0, \) and \( \alpha \)
We can rewrite the expression for \( T \):

\[ T = 2\pi \cdot m^{1/2} \cdot (U_0)^{-1/2} \cdot (\alpha)^{-1} \]

### Step 5: Identify the Exponents
From the equation \( T \propto m^{a} U_0^{b} \alpha^{c} \), we can identify:

- The exponent of \( m \) is \( a = \frac{1}{2} \)
- The exponent of \( U_0 \) is \( b = -\frac{1}{2} \)
- The exponent of \( \alpha \) is \( c = -1 \)

### Step 6: Calculate the Product \( abc \)
Now we calculate the product \( abc \):

\[ abc = \left(\frac{1}{2}\right) \cdot \left(-\frac{1}{2}\right) \cdot (-1) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \]

### Final Answer
Thus, the product \( abc \) is:

\[ \boxed{\frac{1}{4}} \]