Answer:
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.
