Simple Harmonic Oscillation with given equation

[kgal]

Homework Statement

A massive object of m = 5.2 kg oscillates with simple harmonic motion. Its position as a function of time varies according to the equation x(t) = 1.6sin(∏t/1.6 + ∏/6).
a. What is the position, velocity and acceleration of the object at t = 0s?
b. What is the kinetic energy of the object as a function of time?
c. At what time after t=0s is the kinetic energy first at a maximum?

Homework Equations

x(t) = 1.6sin(∏t/1.6 + ∏/6).

The Attempt at a Solution

a. x(0) = 1.6sin(0 + ∏/6) = 0.8 m
v(0) = ∏cos[(∏t / 1.6) + (∏/6)] = ∏cos(∏/6) = 2.72 m/s
a(0) = - (∏^2 / 1.6) sin [(∏t / 1.6) + (∏/6)] = - 5.33 m/s^2 (the right answer should be -3.08 m/s^2)!

b. K(t) = 1/2mv(t)^2 = 1/2m{[∏cos[(∏t / 1.6) + (∏/6)]}^2 //plugged v from part a
2.6∏^2cos^2[(∏t / 1.6) + (∏/6)]

c. K' = -100sin[(∏t / 1.6) + (∏/6)]cos[(∏t / 1.6) + (∏/6)].
K'(t) = 0 at t = .26 s (But the right answer should be 1.33 s)!

 

[Curious3141]

Homework Statement

A massive object of m = 5.2 kg oscillates with simple harmonic motion. Its position as a function of time varies according to the equation x(t) = 1.6sin(∏t/1.6 + ∏/6).
a. What is the position, velocity and acceleration of the object at t = 0s?
b. What is the kinetic energy of the object as a function of time?
c. At what time after t=0s is the kinetic energy first at a maximum?

Homework Equations

x(t) = 1.6sin(∏t/1.6 + ∏/6).

The Attempt at a Solution

a. x(0) = 1.6sin(0 + ∏/6) = 0.8 m
v(0) = ∏cos[(∏t / 1.6) + (∏/6)] = ∏cos(∏/6) = 2.72 m/s
a(0) = - (∏^2 / 1.6) sin [(∏t / 1.6) + (∏/6)] = - 5.33 m/s^2 (the right answer should be -3.08 m/s^2)!

Remember that sin(π/6) = 1/2 NOT √3/2!

b. K(t) = 1/2mv(t)^2 = 1/2m{[∏cos[(∏t / 1.6) + (∏/6)]}^2 //plugged v from part a
2.6∏^2cos^2[(∏t / 1.6) + (∏/6)]

c. K' = -100sin[(∏t / 1.6) + (∏/6)]cos[(∏t / 1.6) + (∏/6)].
K'(t) = 0 at t = .26 s (But the right answer should be 1.33 s)!

Homework Statement

For part c), it's much easier to use a double-angle formula for cosine to express the KE as a power-one expression of a single cosine term plus a constant, then use the periodicity of the cosine function to find where the KE is next at a maximum.

I didn't check your differentiation thoroughly, but remember that if you use differentiation and set the derivative to 0 (in which case either that sine term is zero OR the cosine term is zero), you get both local minima and local maxima, so you need to figure out which is which. A little tedious.

 

[kgal]

*for b. Duh! thanks :)

*for c. I set 3pi/2 = pit/1.6 + pi/6 (cos function) and got t = 2.13 s, which is larger than the t i got when i set pit/1.6 + pi/6 = 0 (sin function), but I still somehow missed the target t = 1.33 s, just don't know where i went wrong...

 

[Curious3141]

*for b. Duh! thanks :)

*for c. I set 3pi/2 = pit/1.6 + pi/6 (cos function) and got t = 2.13 s, which is larger than the t i got when i set pit/1.6 + pi/6 = 0 (sin function), but I still somehow missed the target t = 1.33 s, just don't know where i went wrong...

Remember that sin 0 = sin π = sin 2π = 0.

When you use the '0' you get a negative time. So try the next one.

By right you should consider the cosine part too. In that case, cos π/2 = cos 3π/2 = 0.

But the sine term holds the answer here.

You really have to check to make sure the KE is a minimum at that point. That involves either calculating the second derivative, or curve sketching. Tedious, tedious.

This is why I suggested an alternative method which really is a lot easier.

 

[kgal]

Thanks alot