GATE 2020 PH – Question 49
A mass m has potential $U(x)=a\sin(k^2x-\pi/2)$, with $a>0,k^2>0$. The small-oscillation angular frequency about x = 0 is
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Correct answer: (B) $k^2\sqrt{a/m}$
Explanation
Rewrite the potential as $U=-a\cos(k^2x)$. Near zero, $U=-a+ak^4x^2/2+\cdots$. Comparing the quadratic term with $m\omega^2x^2/2$ gives $m\omega^2=ak^4$, hence $\omega=k^2\sqrt{a/m}$, B. The argument contains $k^2x$, so the second derivative produces $k^4$.
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