The Kinetic energy of a simple hormonic oscillating with angular frequency of 176 rad/s. The frequency of the simple harmonic oscillator is −Hz(π=22/7)

The correct option is 88.

Step-by-Step Explanation:

In a simple harmonic motion (SHM), the displacement of the oscillator as a function of time $t$ is given by:
$x\left(t\right)=A \sin (\omega t+\phi )$
where $\omega$ is the angular frequency of the simple harmonic oscillator.

The velocity of the oscillator is:
$v\left(t\right)=\frac{dx}{dt}=A\omega \cos (\omega t+\phi )$

The kinetic energy ($KE$) of the oscillator is given by:
$KE=\frac{1}{2}m{v}^2=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2(\omega t+\phi )$

Using the trigonometric identity ${\cos }^2\theta =\frac{1+\cos (2\theta )}{2}$, we can rewrite the kinetic energy as:
$KE=\frac{1}{4}m{A}^2{\omega }^2[1+\cos (2\omega t+2\phi \left)\right]$

This shows that the kinetic energy oscillates at twice the angular frequency of the simple harmonic oscillator:
${\omega }_{KE}=2\omega$

We are given that the kinetic energy oscillates with an angular frequency of:
${\omega }_{KE}=176\text{ rad/s}$

Substituting this value into the relationship:
$176=2\omega$

Solving for the angular frequency of the simple harmonic oscillator ($\omega$):
$\omega =\frac{176}{2}=88\text{ rad/s}$

Thus, the corresponding value for the oscillator's angular frequency is 88.