An air-track glider is attached to a spring. the glider is pulled to the right and released from rest at =0 s. it then oscillates with a period of 1.50s and a maximum speed of 50.0 cm/s .

what is the amplitude of the oscillation?

what is the glider's position a t = 29.0s?

a) = 0.12 m b) 0.06 m to the left of the equilibrium point.

Explanation:

Applying the energy conservation principle, we can write :

1/2*k*A² = 1/2*k*(Δx)² + 1/2*m*v²

When the speed is maximum, the block is crossing through the equilibrium point, so all the energy is kinetic, and we can write:

1/2*k*A² = 1/2*m*vmax² (1)

As the block and the spring are oscillating in a SHM, we can apply the definition of angular frequency, as follows:

ω² = k/m

We can convert the angular frequency to linear frequency as follows:

ω = 2*π*f

The frequency is just the number of oscillations per unit time, so, is the inverse of the period:

f = 1/T ⇒ ω = 2*π / T

⇒ (2*π/T)² = k/m

Replacing in (1), and rearranging terms, we can solve for A:

A = vmax* / ω = vmax / 2*π*f  ⇒ vmax * T / 2*π* = 0.12 m

b) As the glider is moving in a SHM, we can find his position at any time, using the following equation:

x(t) = A cos (ωt +φ)

As we know that at t= 0 s, the block is released from rest, and that at t=0, the displacement will be equal to the amplitude of the SHM, this means that we will have the following equations:

x(t) = B cos (ωt + φ)

dx/dt = -ωB sin (ωt + φ)

At t=0, we have x (0) = A and dx/dt (0) = 0, so B= A and φ=0º

⇒ x(t)  = A cos ωt

For t= 29.0 s, we have:

x(29) = 0.12 m cos ((2π/1.5 s)*29.0) sec = -0.06 m

This means that it will be 0.06 m to the left of the equilibrium point.