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

`t = (5)/(12) s`

Step-by-step 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

`(1)/(2) mv^2 = (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 = (\pi)/(4) or (3\pi)/(4)`

`\pi t = (3\pi)/(4) - (\pi)/(3)`

`\pi t = (5\pi)/(12)`

`t = (5)/(12) s`