The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 text rad/s$176 \text{ rad/s}$. The frequency of this simple harmonic oscillator is ________ Hz. [texttake pi = frac227]$[\text{take } \pi = \frac{22}{7}]$

### Related Formula omega_k = 2pi f_k$$\omega_k = 2\pi f_k$$ f_textoscillator = fracf_k2$$f_{\text{oscillator}} = \frac{f_k}{2}$$ ### Core Logic For a simple harmonic oscillator with an angular frequency omega$\omega$, the kinetic energy oscillates at double the angular frequency (2omega$2\omega$). The given angular frequency of kinetic energy oscillation is omega_k = 176 text rad/s$\omega_k = 176 \text{ rad/s}$. ### Step 1: Finding Kinetic Energy Frequency First, find the frequency of kinetic energy oscillations (f_k$f_k$): f_k = fracomega_k2pi = frac1762 times frac227$$f_k = \frac{\omega_k}{2\pi} = \frac{176}{2 \times \frac{22}{7}}$$ f_k = frac176 times 744 = 4 times 7 = 28 text Hz$$f_k = \frac{176 \times 7}{44} = 4 \times 7 = 28 \text{ Hz}$$ ### Step 2: Final Conclusion Since kinetic energy oscillates at twice the frequency of the oscillator itself: f_textoscillator = fracf_k2 = frac282 = 14 text Hz$$f_{\text{oscillator}} = \frac{f_k}{2} = \frac{28}{2} = 14 \text{ Hz}$$ ### Pattern Recognition KE and PE always oscillate at twice the frequency (2nu$2\nu$) of the underlying SHM (v$v$) due to the sin^2(omega t)$\sin^2(\omega t)$ or cos^2(omega t)$\cos^2(\omega t)$ terms resolving to (1 - cos(2omega t))/2$(1 - \cos(2\omega t))/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations