Question

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?

Answer 1

Answer:

$t = \frac{5}{12} s$

Explanation:

As we know that the equation of particle position is given as

$x = A cos(\omega t + \phi_0)$

Now the speed of the particle is given as

$v = A\omega sin(\omega t + \phi_0)$

now we know that potential energy and kinetic energy is equal

so we have

$\frac{1}{2} mv^2 = \frac{1}{2}kx^2$

so we will have

$A^2\omega^2 sin^2(\omega t + \phi_0) = A^2\omega^2 cos^2(\omega t + \phi_0)$

$tan^2(\omega t + \phi_0) = 1$

$\omega t + \phi_0 = \frac{\pi}{4} or \frac{3\pi}{4}$

$\pi t = \frac{3\pi}{4} - \frac{\pi}{3}$

$\pi t = \frac{5\pi}{12}$

$t = \frac{5}{12} s$