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Question

(A) The standard equation for velocity in simple harmonic motion $(SHM)$ is $v^2 = \omega^2 (A^2 - x^2)$, where $\omega$ is the angular frequency and

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(A) The standard equation for velocity in simple harmonic motion $(SHM)$ is $v^2 = \omega^2 (A^2 - x^2)$, where $\omega$ is the angular frequency and $A$ is the amplitude.Given the equation $v^2 = 50 - x^2$, we can rewrite it as $v^2 = 1(50 - x^2)$.Comparing this with the standard equation, we get $\omega^2 = 1$, which implies $\omega = 1 \ 	ext{rad/s}$.The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{1} = 2\pi \ 	ext{s}$.Using the approximation $\pi \approx \frac{22}{7}$, we get $T = 2 	imes \frac{22}{7} = \frac{44}{7} \ 	ext{s}$.According to the problem, $T = \frac{x}{7} \ 	ext{s}$.Equating the two expressions for $T$: $\frac{x}{7} = \frac{44}{7}$.Therefore, $x = 44$.

The velocity of a particle executing simple harmonic motion along the $x$-axis is described by the equation $v^2 = 50 - x^2$, where $x$ represents displacement. If the time period of the motion is $\frac{x}{7} \ \text{s}$, the value of $x$ is . . . . . . .

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