The kinetic energy of a simple harmonic oscillator is oscillating...

In a simple harmonic oscillator, if the displacement is $x = A \sin(\omega t + \phi)$, the velocity is $v = A\omega \cos(\omega t + \phi)$.
The kinetic energy is given by $K = \frac{1}{2}mv^2 = \frac{1}{2}m A^2 \omega^2 \cos^2(\omega t + \phi)$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$, we get $K = \frac{1}{4}m A^2 \omega^2 [1 + \cos(2\omega t + 2\phi)]$.
The angular frequency of the kinetic energy oscillation is $\omega_{KE} = 2\omega$, where $\omega$ is the angular frequency of the SHM.
Given $\omega_{KE} = 176$ rad/s, we have $2\omega = 176$, which gives $\omega = 88$ rad/s.
The frequency of the simple harmonic oscillator is $f = \frac{\omega}{2\pi}$.
Substituting the values: $f = \frac{88}{2 \times \frac{22}{7}} = \frac{88 \times 7}{44} = 2 \times 7 = 14$ Hz.