Answer:
[tex]t = \frac{5}{12} s[/tex]
Explanation:
As we know that the equation of particle position is given as
[tex]x = A cos(\omega t + \phi_0)[/tex]
Now the speed of the particle is given as
[tex]v = A\omega sin(\omega t + \phi_0)[/tex]
now we know that potential energy and kinetic energy is equal
so we have
[tex]\frac{1}{2} mv^2 = \frac{1}{2}kx^2[/tex]
so we will have
[tex]A^2\omega^2 sin^2(\omega t + \phi_0) = A^2\omega^2 cos^2(\omega t + \phi_0)[/tex]
[tex]tan^2(\omega t + \phi_0) = 1[/tex]
[tex]\omega t + \phi_0 = \frac{\pi}{4} or \frac{3\pi}{4}[/tex]
[tex]\pi t = \frac{3\pi}{4} - \frac{\pi}{3}[/tex]
[tex]\pi t = \frac{5\pi}{12}[/tex]
[tex]t = \frac{5}{12} s[/tex]