An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t=0s.?

Question

An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at =0s. It then oscillates with a period of 1.50s and a maximum speed of 36.0cm/s.

A)What is the amplitude of the oscillation?

B)What is the glider's position at = 20.0s?

Answer 1

T = period = 1.50 s

V = maximum speed = 0.360 m/s



The angular frequency is given by:

ω = 2π / T = 4π/3 rad/s



The amplitude is given by:

A = V / ω = 0.27/π m

A = 0.0859 m = 8.59 cm < - - - - - - - - - - - - - - - - - - - - - - - - answer A



The equation for position as a function of time is:

x = A cos( ω t )

x = (0.27/π m) cos( (4π/3 rad/s) t )



At time t = 20 s, the position is:

x = (0.27/π m) cos( (4π/3 rad/s) (20 s) )

x = -0.135/π m = -A/2

x = -4.30 cm < - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - answer B

Answer 2

Air Track Glider

Answer 3

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The period of SHM is given by: T = 2π ∙ √(m/k) so T / 2π = √(m/k) You know that the maximum elastic potential energy (when compression or stretching is equal to the amplitude and speed is 0) must be the same as the maximum kinetic energy (when compression or stretching is 0 and speed is at maximum), so k A² / 2 = m V² / 2 k A² = m V² A² = V² m / k A = V √(m/k) A = V T / 2π A = (0.6 m/s) (1.5 s) / 2π A = 0.1432 m < - - - - - - - - - - - - - answer The general equation for position of SHM is: x = A cos(ωt + φ) where x = position at time t A = amplitude ω = angular velocity t = time φ = phase As you are releasing the system at time t=0, the phase φ = 0 rad. The angular velocity can be found by: ω = 2π / T ω = (2π rad) / (1.5 s) ω = 4.189 rad/s So the equation for position is: x = (0.1432 m) cos((4.189 rad/s)t) At time t=25 s, you get x = (0.1432 m) cos((4.189 rad/s)(25)) x = -0.071 m < - - - - - - - - - - - - - - - - - - - answer (Note: you must put your calculation in RAD modus)