Question

At what displacment velocity of body is 1 over 2 under root

Answer 1

$\mathbb{\underline {ANSWER}}$

At displacement ±A/√2

$\mathbb{\underline {EXPLANATION}}$

Maximum velocity = Aω   where, A is the amplitude

                                                      ω is the angular frequency

Velocity given = $\frac{1}{\sqrt{2} } V_{max} = \frac{A\omega}{\sqrt{2} }$

We know that,

Velocity of a body in SHM at any point can be given by the equation

$\boxed{V =\omega\sqrt{A^{2}-x^{2} } }$

where x is the displacement from mean position

Substituting the values we get,

$\frac{A\omega}{\sqrt{2} } = \omega\sqrt{A^{2}- x^{2} }$

Cancelling ω on both sides we get,

$\frac{A}{\sqrt{2} } = \sqrt{A^{2}-x^{2} }$

Squaring on both sides,

$\frac{A^{2} }{2} = A^{2}- x^{2}$

$x^{2} = A^{2} -\frac{A^{2} }{2}$

$x^{2} = \frac{A^{2} }{2}$

$x = \pm\frac{A}{\sqrt{2} }$