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

Motion Of A Particle

EDIT – added alternative solution at the bottom.
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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
ω = 13 (rad/s)
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v = dx/dt = Aωsin(ωt)
Eₚ = ½kx²
Eₖ =½mv²
We want Eₖ = 2Eₚ
½mv² = 2(½kx²)
mv² = 2kx²
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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
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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.