Question

A 0.750 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.900 N to displace the glider to a new equilibrium position, x= 0.090 m.

a) Find the effective spring constant of the system.

b) The glider is now released from rest at x= 0.090 m. Find the maximum x-acceleration of the glider.

c) Find the x-coordinate of the glider at time t= 0.550T, where T is the period of the oscillation.

d) Find the kinetic energy of the glider at x=0.00 m.

Answer 1

mass m -0.750 kg the horizontal applied force F-0.9 N the di splacement of new equilibrium position is x = 0.09m the restoring force F --Kx the effective spring constant K =- 0.9 0.09 -10 Nim time t=0.550 T the angular frequency =- m V0.75 the mazimum accelerati on is a =-ar x cos at (37)2(0.09)cos- 0.550 T =1 1978 m/s 2 X=Acos at where Als amplitude-0.11m =(0.09) cos|-× 0.550 T =-0.0898 m kinetic energy at x -0.00m v=A ω= (0.09) (3.7) = 0.328 m/s 0.750)(0.328)2 =001518J K =

(or)

A) k_eff = F/x = 0.9/0.09

    = 10 N/m

B) Maximum acceleration = Aω²

where A=Amplitude = 0.09 m , ω=√(m/k) = 0.27386 rad/s^{2 ,} ω² = 0.075

Maximum acceleration = 0.00675 m/s² = 6.75 cm/s²

C.

X=Acos(ωt)=Acos(2πt/T)

At t=0.55T,

x=0.09cos( 2π x 0.55) = -0.0898 m

D.

K=(1/2) mv²=(1/2) m(Aω)²

                     = 2.28x10^-4 J