Answer: \(\displaystyle \frac{\omega_1}{\omega_2}=\frac{\sqrt3}{2}\).

For a particle moving in a horizontal circle on the smooth inner surface, let the normal reaction be \(N\) and let the angle between the upward vertical and the reaction be \(\theta\).

Resolving vertically, there is no vertical acceleration, so

\(N\cos\theta=mg.\)

Resolving horizontally towards the centre of the horizontal circle, the radius of the horizontal circle is \(r\sin\theta\). Hence

\(N\sin\theta=m(r\sin\theta)\omega^2.\)

Since \(\sin\theta\ne0\), this gives

\(N=mr\omega^2.\)

Combining with \(N\cos\theta=mg\),

\(mr\omega^2\cos\theta=mg, \qquad r\omega^2=\frac{g}{\cos\theta}.\)

For the first case, \(\tan\theta_1=\frac34\), so \(\cos\theta_1=\frac45\). Therefore

\(r\omega_1^2=\frac{g}{4/5}=\frac54g.\)

For the second case, \(\tan\theta_2=\frac43\), so \(\cos\theta_2=\frac35\). Therefore

\(r\omega_2^2=\frac{g}{3/5}=\frac53g.\)

Divide the two equations:

\(\frac{\omega_1^2}{\omega_2^2} =\frac{\frac54g}{\frac53g} =\frac34.\)

Since angular speeds are positive,

\(\frac{\omega_1}{\omega_2}=\frac{\sqrt3}{2}.\)