# The motion of a particle is given by x(t)=(25cm)cos(13t), where t is in s.?

0

find the first time at which the kinetic energy is twice the potential energy.

• Motion Of A Particle

• EDIT – added alternative solution at the bottom.

___________________________

If you are allowed to use standard equations, just consider a mass (m) on a spring (spring constant k):

x = Acos(ωt) where ω = √(k/m)

Note, since x(t)=(25cm)cos(13t), we can match coefficients so:

A = 25cm (=0.25m) and

________________________________________

v = dx/dt = -Aωsin(ωt)

Eₚ = ½kx²

Eₖ =½mv²

We want Eₖ = 2Eₚ

½mv² = 2(½kx²)

mv² = 2kx²

________________________________________

Since x = Acos(ωt) and v = -Aωsin(ωt)

m(-Aωsin(ωt))² = 2k(Acos(ωt))²

mA²ω²sin²(ωt) = 2kA²cos²(ωt)

mω²sin²(ωt) = 2kcos²(ωt)

sin²(ωt)/cos²(ωt) = 2k/(mω²)

ω = √(k/m) so ω² = (k/m), and sin()/cos() = tan() so;

tan²(ωt) = 2k/(m(k/m)

tan²(ωt) = 2

tan(ωt) = √2

ωt = tan⁻¹√2

Remember to switch calculator to radians:

ωt = 0.9553

t = 0.9553/ω = 0.9553/13 = 0.0735s

_____________________________________________

Alternatively, you can find where the kinetic energy is 2/3 of its maximum value because then 1/3 of the total energy is potential energy.

v max (when sn()=1) =

KE = ½mv² = ½m(-Aωsin(ωt))² = ½mAω²sin²(ωt)

The max. value odof sin() is 1, so the max value of KE = is when sin() = 1:

KEmax = ½mAω²1² = ½mAω²

When KE = (2/3)KEmax

½mAω²sin²(ωt) = (2/3)½mAω²

ω²sin²(ωt) = (2/3)ω²

sin²(ωt) = (2/3)

sin(ωt) = √(2/3)

ωt = sin⁻¹√(2/3)

. . .= 0.9553

Then the working is the same as in the first method.

Also Check This  how did lavoisier transform the field of chemistry in the late 1700’s?