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  4. A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm / s. The distance of the particle from the mean position when its speed becomes 5 cm / s is:

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Q. A particle executes simple harmonic motion with an amplitude of $4 cm$. At the mean position the velocity of the particle is $10 cm / s$. The distance of the particle from the mean position when its speed becomes $5 cm / s$ is:

Oscillations

Solution:

$A \omega=10 \& A=4$
so $\omega= \frac{5}{2}$
$v=\omega \sqrt{A^{2}-x^{2}}=2 cm / s$