The displacement of two identical particles performing S.H.M are...

The total energy of a particle performing simple harmonic motion is given by $E = \dfrac{1}{2} m \omega^2 A^2$

For the first particle, the amplitude is $A_1 = 6$ and the angular frequency is $\omega_1 = 5$.

Its energy is $E_1 = \dfrac{1}{2} m (5)^2 (6)^2 = \dfrac{1}{2} m (25)(36) = 450 m$

For the second particle, the amplitude is $A_2 = 4$ and the angular frequency is $\omega_2 = \omega$.

Its energy is $E_2 = \dfrac{1}{2} m (\omega)^2 (4)^2 = 8 m \omega^2$

Since the particles are identical, their masses are the same. Given that their energies are equal, we have $E_1 = E_2$

$450 m = 8 m \omega^2$

$\omega^2 = \dfrac{450}{8} = \dfrac{225}{4}$

$\omega = \dfrac{15}{2} = 7.5$ unit

Answer: $7.5$ unit