Question

The displacement of a block of mass 1.280 kg attached to a spring whose spring constant is 50 N/m is given by x = A cos ωt, where A = 12 cm. In the first complete cycle, find the values of x and t at which the kinetic energy is equal to one half the potential energy.

Answer 1

The KE is half the PE when the KE is 1/3 of the TME(total mechanical energy) of the system.

TME = max KE = ½kA² = ½ x 50N/m x (0.12m)² = 0.36 J

We're interested in where KE = TME / 3 = 0.12 J

and

0.12 J = ½mv² = ½ x 1.280kg x v²,

or v = ±0.43.3 m/s = ± 43.3 cm/s

v(t) = x'(t) = Aωcos(ωt)

and ω = √(k/m) = √(50kg/s² / 1.280kg) = 6.25 rad/s, so

v(t) = 0.12m x 6.25rad/s x cos(6.25t) = 0.75m/s x cos(6.25t)

Therefore,

cos(6.25t) = ±0.433m/s / 0.75m/s = ±0.577

6.25t = arccos0.577 = ±0.96 rads

→ t = ± 0.156 s

6.25t = arccos-0.577 = ±2.19 rads

→ t = ± 0..350 s

For the "negative" times, we have to subtract those values from one full period, where period T = 2π/ω = 1.005 s. Therefore

time 1 = 0.156 s

time 2 = 0.350 s

time 3 = 1.005 s - 0.156s = 0.849 s

time 4 = 1.005 s - 0.350s = 0..655 s

Plugging this values in the formula for getting positions x(t) = 12cm x sin(6.25t)

the values of x and t at which the kinetic energy is equal to one half the potential energy:

At t=0.156 ; x(t) = 12cm x sin(6.25x0.156) = 0.204 cm

at t = 0.350s ; x(t) = 12cm x sin(6.25x0.350) = 0.458 cm

at t = .849; x(t) = 12cm x sin(6.25x0.849) = 1.11cm

at t =.655 s; x(t) = 12cm x sin(6.25x0.655) =.856cm

Hope this will help you. Thanks.