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$V = A \omega \sin(\omega t + \theta)$ gives velocity of a particle in SHM at time $t$. But, why does the value of $\omega$ doesn't change when $V$ is changed?

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    $\begingroup$ Why should it? . $\endgroup$ Jan 26, 2019 at 17:18

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For simple harmonic motion, $\omega$ is a constant of the motion and is equal to $\dfrac{2\pi}{\rm period}$

For example, for a spring-mass system $\omega = \sqrt{\frac k m}$ where $k$ is the spring constant and $m$ a mass (which is constant) at the end of the spring.

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$\omega$ in this case is the angular frequency, not an angular velocity. It is given this term because in the equation for the position of the particle is given $x(t)=A\sin(\omega t)$ the input to the $\sin$ function is an angle $\omega t$. It just tells you how "fast" the angle changes over time.

Typically the angular frequency just depends on what forces are acting on the object, and in many cases (like a mass on an ideal spring or a simple pendulum) the forces do not depend on velocity, so $\omega$ doesn't either. In cases where forces do depend on velocity (like damped oscillators) things are more complicated, but the natural frequency of the system when damping is not present still plays an important role in describing the motion of the system over time.

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