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

Part A

Write down the equation giving its position y (+ upward) as a function of time t. Assume the object started by being compressed 17cm from the equilibrium position (where y = 0), and released.

Write down the equation giving its position ( upward) as a function of time . Assume the object started by being compressed 17 from the equilibrium position (where = 0), and released.

y(t)=(0.17m)?cos(2?t0.65s)
y(t)=(0.17m)?sin(2?t0.65s)
y(t)=(0.17m)?cos(0.65s?t)
y(t)=(0.17m)?cos(t0.65s)

Part B

How long will it take to get to the equilibrium position for the first time?

Express your answer to two significant figures and include the appropriate units.

Part C

What will be its maximum speed?

Express your answer to two significant figures and include the appropriate units.

Part D

What will be the object's maximum acceleration?

Express your answer to two significant figures and include the appropriate units.

Part E

Where will the object's maximum acceleration first be attained?

Where will the object's maximum acceleration first be attained?

a.release point

b,equilibrium point

Respuesta :

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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

A) y(t) = (0.17 m)*cos((2π/0,65)*t)

(B) It will take , t = 0.1625 sec

(C) Its maximum speed,  Vmax = 1,64 m/s

(D) Its maximum acceleration,  Amax = 15,88 m/s²

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

Solving each part one by one:

(A)

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:

y(t) = (0.17 m)*cos((2π/0,65)*t)

(B) This means, finding 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) 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) 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)

For the acceleration, ocurrs 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, ocurrs 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

Find more information about Acceleration here:

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