Physics, asked by sreekrishna37, 10 months ago

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

Answers

Answered by aristocles
0

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

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Topic : SHM

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