Question

A 140 g block hangs from a spring with spring constant 16 N/m . At t=0s the block is 28 cm below the equilibrium point and moving upward with a speed of 110 cm/s .

I did find the oscillation frequency it was 1.70Hz

a)What is the block's distance from equilibrium when the speed is 63 cm/s ?

b)What is the block's distance from equilibrium at t=11s?

Answer 1

ω = sqrt(k/m) = sqrt(16N/m / 0.14kg) = 10.690 rad/s ← frequency in radians/s
f = ω/2π = 1.70Hz ← frequency in Hz, which is what they want, I think: #1

At t = 0, system energy E = Ep + Ek
E = ½kx² + ½mv² = ½*16N/m*(0.28m)² + ½*0.14kg*(110m/s)² = 0.711 J
When the speed is 0.63m/s,
Ek = ½mv² = ½*0.14kg*(0.63m/s)² = 0.0277 J
So Ep = E - Ek = 0.711J - 0.0277J = 0.683J = ½kx² = ½*16N/m*x²
x = 0.292m = 29.2 cm ← #2

3. Well, the amplitude is found from
E = Ep = 0.711J = ½kx² = ½*16N/m*x²
x = 0.292 m = 29.2 cm
Using the general form
x(t) = Acos(ωt - φ)
and plugging in our initial conditions:
28 = 29.2*cos(0 - φ)
28/29.2= 0.958 = cos(-φ)
φ = - arccos0.958 =-16.664 = -0.290 rad
So x(t) = 29.2cm * cos(ωt + 0.290)
When t = 11s,
x(1) = 29.2cm * cos(10.690*1 + 0.290) = 28.665 cm,