Solution: Small Oscillations in a Potential Field
The potential energy is given by: $$ U(x) = -k(x-a)^2(x-b)^2 $$ We need to find the point of stable equilibrium. A system oscillates about a minimum in the potential energy curve (where \( U \) is lowest).
At \( x=a \) and \( x=b \), \( U=0 \). For any other \( x \), the term \((x-a)^2(x-b)^2\) is positive, so \( -k(…) \) is negative. This means the graph of \( U(x) \) goes down from 0 at \( a \) and \( b \) to a negative minimum in between. This implies there is a single stable equilibrium position exactly in the middle of \( a \) and \( b \).
$$ x_0 = \frac{a+b}{2} $$To find the frequency, let’s substitute a new variable \( y \) that represents the displacement from the center \( x_0 \). Let \( x = x_0 + y = \frac{a+b}{2} + y \).
Also, let \( L = \frac{b-a}{2} \) be the half-distance between \( a \) and \( b \). Then: $$ x – a = y + L $$ $$ x – b = y – L $$
Substitute these into the potential energy equation: $$ U(y) = -k [ (y+L)(y-L) ]^2 $$ $$ U(y) = -k [ y^2 – L^2 ]^2 $$
Expand the squared term: $$ U(y) = -k [ y^4 – 2y^2 L^2 + L^4 ] $$ $$ U(y) = -ky^4 + 2kL^2 y^2 – kL^4 $$
For small oscillations, \( y \) is very small, so \( y^4 \) is negligible compared to \( y^2 \). We can ignore the \( y^4 \) term. The constant term \( -kL^4 \) affects the energy level but not the force/frequency. The effective potential is: $$ U_{eff}(y) \approx \frac{1}{2} (4kL^2) y^2 + \text{const} $$
Comparing this to the standard SHM potential \( U = \frac{1}{2} K_{eff} y^2 \), we identify the effective spring constant: $$ K_{eff} = 4kL^2 $$
Substitute \( L = \frac{b-a}{2} \) back into the expression for \( K_{eff} \): $$ K_{eff} = 4k \left( \frac{b-a}{2} \right)^2 = 4k \frac{(b-a)^2}{4} = k(b-a)^2 $$
The angular frequency \( \omega \) is: $$ \omega = \sqrt{\frac{K_{eff}}{m}} = \sqrt{\frac{k(b-a)^2}{m}} $$
Conclusion: The particle oscillates about a single location with angular frequency \( (b-a)\sqrt{k/m} \). This matches option (c).