The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/ s. The frequency of this simple harmonic oscillator is _____Hz. [take $$\pi = \frac{22}{7}$$]

We are told that the kinetic energy of a simple harmonic oscillator oscillates with an angular frequency of 176 rad/s and we need to find the frequency of the SHM oscillator itself.

For a particle executing SHM with displacement $$x = A\sin(\omega_0 t)$$, where $$\omega_0$$ is the angular frequency of the SHM and $$A$$ is the amplitude.

Differentiating with respect to time gives the velocity: $$v = A\omega_0\cos(\omega_0 t)$$

This leads to the kinetic energy: $$ KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}mA^2\omega_0^2\cos^2(\omega_0 t) $$

Using the identity $$\cos^2(\theta) = \tfrac{1 + \cos(2\theta)}{2}$$, we obtain:

$$ KE = \tfrac{1}{2}mA^2\omega_0^2 \times \tfrac{1 + \cos(2\omega_0 t)}{2} = \tfrac{mA^2\omega_0^2}{4}\left(1 + \cos(2\omega_0 t)\right) $$

It follows that the kinetic energy has a constant part and an oscillating part $$\cos(2\omega_0 t)$$, so its angular frequency is $$2\omega_0$$, which is twice the angular frequency of the SHM.

Since the KE oscillates at $$\omega_{KE} = 176$$ rad/s, we set:

$$ 2\omega_0 = 176 $$

$$ \omega_0 = \frac{176}{2} = 88 \text{ rad/s} $$

Using $$\omega_0 = 2\pi f$$, we have:

$$ f = \frac{\omega_0}{2\pi} $$

Substituting $$\pi = \frac{22}{7}$$ yields:

$$ f = \frac{88}{2 \times \frac{22}{7}} = \frac{88}{\frac{44}{7}} = \frac{88 \times 7}{44} = \frac{616}{44} = 14 \text{ Hz} $$

The frequency of the simple harmonic oscillator is 14 Hz.

The correct answer is Option (4): 14.