A particle oscillates harmonically x = A cos(ωt + φ0), with amplitude 9 m, angular frequency π s −1, and initial phase π 3 radians. Every now and then, the particle’s kinetic energy and potential energy happen to be equal to each other (K = U). When does this equality happen for the first time after t = 0?

Respuesta :

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]

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