# §☺☻Oscillations and Kinetic Energy... is for exam!!!?

A 0.50-kg object is attached to an ideal massless spring of spring constant 20 N/m along ahorizontal, frictionless surface. The object oscillates in simple harmonic motion and has aspeed of 1.5 m/s at the equilibrium position

I worked out the amplitude, is 1.5/2sqrt(10

) ~~~> 0.237m

◘ At what location are the kinetic energy and the potential energy of the system the same?

Relevance

m = 0.5 kg

k = 20 N/m

w = √k/m = √20/0.5 = 6.32455532

max speed = V = 1.5 = Aw = 6.32455532A

A = 0.237 m {confidence check :>}

max mech. energy = KE at equilibrium position = 1/2mV² = 0.5625 J

Let where KE becomes 1/2 its value or = 0.28125 J be called x.

SPE = 0.28125 = 1/2kx² = (0.5)(20)x² = 10x²

x² = 0.028125

x ≈ 0.168 m ANS*

*Clarification:

Total Mechanical Energy = Σ (KE + U) = 0.5625 J

So when KE = 0.28125 J, U must equal the same.

• Miguel5 years agoReport

I got it know, never mind :)

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• We know that the potential energy is a max when V = 0 at the max distance from the equilibrium location where x = A and is zero at the equilibrium where V is a max where x = 0

PE = ½kx² at x = A PE is at the maximum and KE = 0

KE = ½mV² at x = 0 (equilibrium) V and KE are at a maximum and PE = 0

Thus at x = A PE = KE at x = 0. In other words, ignoring losses, the energy transfers from PE to KE and back

E = ½kA² = ½mV² (at equilibrium where x = 0)

x(t) = Acosωt where ω = √k/m = √20/0.5) = √40

Vmax = ωA = 1.5 => A =1.5/√40 = 0.237

PEmax = ½*k*A² = ½*20*9/160 = 9/16 = ½*0.5*1.5² = 9/16

We know that ½kx² + ½mV² = 10x²+V²/4 = 9/16 -----> 40x² + V² = 9/4 -----> V² = 9/4 - 40x²

Now set ½mV² = ½kx² -----> V²/4 = 10x² ----> V² = 40x² = 9/4 - 40x² ----> 80x² = 9/4

x² = 9/320 -----> x = 3/[8*√5] = 3√5/40 = 0.168m <-----------

Check PE = 10x² = 0.28125 J

KE = 9/16 - 0.28125 = 0.28125 J so it checks

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