Question

using definition of linear shm derive an expression for angular frequency of a body in shm​

Answer 1

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Answer 2

The expression for angular velocity of a body in SHM is   $\omega = \sqrt{\frac{k}{m} }$ .

Explanation :

We know that the condition for  a body to be in SHM is that there should be a restoring force :

$F = - kx$  

Where x is displacement from mean position and k is restoring force constant .

∴F = mass $\times$ acceleration = ma

So , ma =  - kx

⇒ $a = \frac{-k}{m}\times x$   - (1)

We know that equation of an SHM is given as :

$x = A sin ( \omega t + \phi)$   -(2) , where $\omega$ is angular frequency

On differentiating it , we get :

$\frac{dx}{dt} = v = A\omega cos(\omega t + \phi)$  -(3)

And on further differentiation we get :

$\frac{d^2x}{dt^2} = a = -A\omega ^2sin (\omega t + \phi)$    -(4)

Now putting the value of x in eq 1 from eq 2 we get :

$a = -\frac{k}{m} A sin (\omega t + \phi)$   -(5)

On comparing eq 1 and eq 5 we get :

$\omega^2 = \frac{k}{m}$

Or, $\omega = \sqrt{\frac{k}{m} }$

So, the expression for angular frequency of a body in SHM is given as :

$\omega = \sqrt{\frac{k}{m} }$