Question

Your body is executing SHM with amplitude a and angular frequency Omega the root mean square speed of the body for one complete oscillation will be position of the particle at equal to zero is mean position

Answer 1

Answer:

it will be at distance $x = \frac{A}{\sqrt2}$ from its mean position when its speed is same as RMS speed

Explanation:

As we know that the speed of SHM is given as

$v = A\omega cos(\omega t + \phi)$

equation of SHM is given as

$x = A sin(\omega t + \phi)$

Here we know that RMS speed for the given expression is

$v_{rms} = \frac{A\omega}{\sqrt2}$

now we know that

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

$\frac{A\omega}{\sqrt2} = \omega \sqrt{A^2 - x^2}$

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

$x = \frac{A}{\sqrt2}$

so it will be at distance $x = \frac{A}{\sqrt2}$ from its mean position

#Learn

Topic : SHM

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