Question

A 0.250 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.

Find the effective spring constant of the system.
1.00×10¹ N/m

The glider is now released from rest at x= 0.090 m. Find the maximum x-acceleration of the glider.
3.60 m/s^2

Find the x-coordinate of the glider at time t= 0.350T, where T is the period of the oscillation.

? - this is where I need help

Find the kinetic energy of the glider at x=0.00 m.
4.05×10^-2 J

Answer 1

1.

F = kx

k = F/x

= (0.900 N) / (0.090 m)

= 10 N/m = 1 x 10¹ N/m

2. a = (k/m)x

Maximum acceleration occurs when x is greatest; in this case, x_max = 0.090 m.

a = (10 N/m)(0.090 m) / (0.250 kg)

= 3.6 m/s^2

3.

x(t) = Acos(ωt + φ), with ω^2 = k/m

A = amplitude = x_max = 0.090 m

φ is a constant; to find it, set t = 0:

t = 0, x(t) = 0.090 m

0.090 m = (0.090 m)(cos(φ))

cos(φ) = 1

φ = 0

T = 2π√ (m/k)

0.350T = 0.350*2π√ (m/k)

t = 0.350*2π√ (m/k)

ωt = 0.350*2π√ (m/k) * √ (k/m)

= 0.70π

x(0.350T) = (0.090 m)cos(0.70π + 0)

= -0.052900672 m