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Question

(D) The speed $v$ of a particle in simple harmonic motion at a displacement $x$ is given by the formula: $v = \omega \sqrt{A^2 - x^2}$, where $\omega$

Vedclass · Accepted Answer

$\frac{A\sqrt{15}}{4}$

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(D) The speed $v$ of a particle in simple harmonic motion at a displacement $x$ is given by the formula: $v = \omega \sqrt{A^2 - x^2}$, where $\omega$ is the angular frequency and $A$ is the amplitude.The maximum speed $v_{max}$ occurs at the mean position $(x = 0)$ and is given by $v_{max} = \omega A$.According to the problem, the speed $v$ is one-fourth of the maximum speed:$v = \frac{v_{max}}{4} = \frac{\omega A}{4}$.Equating the two expressions for $v$:$\omega \sqrt{A^2 - x^2} = \frac{\omega A}{4}$.Dividing both sides by $\omega$:$\sqrt{A^2 - x^2} = \frac{A}{4}$.Squaring both sides:$A^2 - x^2 = \frac{A^2}{16}$.Rearranging to solve for $x^2$:$x^2 = A^2 - \frac{A^2}{16} = \frac{15A^2}{16}$.Taking the square root:$x = \sqrt{\frac{15A^2}{16}} = \frac{A\sqrt{15}}{4}$.Thus, the displacement is $\frac{A\sqrt{15}}{4}$.

$A$ particle executing simple harmonic motion starts from the mean position with amplitude $A$ and periodic time $T$. At what displacement is its speed one-fourth of the maximum speed?

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