Question

A 2.5kg object oscillates at the end of a vertically hanging light spring once every 0.65s .

Answer 1

Answer:

Explanation:

By the general expresion for this problem, we have:

y(t) = A*cos(w*t+∅)

since: w = 2π/T = 2π/0,65  

For the initial conditions:

y(0) = (0.17cm)*cos(w*0+∅) = + 0.17 m ---> cos(∅) = 1 ---> ∅ = 0°

Then:

A) y(t) = (0.17 m)*cos((2π/0,65)*t)

B part

This means, find the first solution for:

y(t) = (0.17 m)*cos((2π/0,65)*t) = 0 > (equilibrium position)

then: cos((2π/0,65)*t) = 0 ---> (2π/0,65)*t = π/2 ---> t = 0.1625 sec

C part

By definition: (Velocity) v = dy/dt

Then, deriving: v = dy/dt = - (0.17 m)*(2π/0,65)*sin((2π/0,65)*t)

The maximum velocity ocurrs when sin((2π/0,65)*t)  = ±1, then (in absolute value): Vmax = 1,64 m/s

D part

By definition: (Aceleration) A = dv/dt

Then, deriving: v = dv/dt = - (0.17 m)*(2π/0,65)²*cos((2π/0,65)*t)

The maximum aceleration ocurrs when cos((2π/0,65)*t)  = ±1, then (in absolute value): Amax = 15,88 m/s²

E part

For the acceleration, ocurres by all the solution when:

cos((2π/0,65)*t)  = cos(phase) = ±1, this means: phase = {π, 2π, 3π, ...}, all π multiples.

Then, for the position:

y(t) = (0.17 m)*±1= {+0.17 m ; -0.17 m}

For the velocity, ocurres by all the solution when:

sin((2π/0,65)*t)  = sin(phase) = ±1, this means: phase = {π/2, π, 3π/2, ...}, all π/2 multiples.

Then, for the position:

y(t) = (0.17 m)*cos(phase)= 0