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 $\pi = \frac{22}{7}$]}
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Energy in SHM (both Kinetic and Potential) oscillates at twice the frequency of the displacement. If the period of SHM is $T$, the period of energy oscillation is $T/2$.
Step 1: Understanding the Concept:
In a simple harmonic oscillator (SHO) with displacement $x(t) = A \sin(\omega t + \phi)$, the kinetic energy is proportional to the square of the velocity. Squaring a trigonometric function doubles its frequency. Thus, the frequency of kinetic energy oscillation is twice the frequency of the displacement oscillation.
Step 2: Key Formula or Approach:
The angular frequency of kinetic energy oscillation ($\omega_{KE}$) is related to the angular frequency of the SHO ($\omega_{SHO}$) by:
\[ \omega_{KE} = 2 \omega_{SHO} \]
The linear frequency $f$ is related to angular frequency by $f = \frac{\omega}{2\pi}$.
Step 3: Detailed Explanation:
Given $\omega_{KE} = 176$ rad/s.
Using the relation $2 \omega_{SHO} = 176$:
\[ \omega_{SHO} = \frac{176}{2} = 88 \text{ rad/s} \]
Now, calculate the linear frequency $f_{SHO}$:
\[ f_{SHO} = \frac{\omega_{SHO}}{2\pi} = \frac{88}{2 \cdot \frac{22}{7}} \]
\[ f_{SHO} = \frac{88 \times 7}{44} = 2 \times 7 = 14 \text{ Hz} \]
Step 4: Final Answer:
The frequency of the simple harmonic oscillator is 14 Hz.
