Question

if the displacement and velocity of a particle executing simple harmonic motion are related through the expression 4v^2= 25-x^2then its time period

Answer 1

$v = \frac{ \sqrt{25 - {x}^{2} } }{2}$

$v \frac{dv}{x} = a$

$a = \frac{ \sqrt{25 - {x}^{2} } }{2} \times \frac{( - 2x)}{ \sqrt{25 - {x}^{2} } } = - x$

$a = - {w}^{2} x = - x$

$w = 1$

$time \: period = \frac{2\pi}{w} = 2\pi$

plz mark brainliest

Answer 2

Answer:

4pi seconds

Explanation:

See the attachment