Simple Harmonic Motion: Finding the Position of a Block at 0.69 seconds

[BlackHayze]

Homework Statement

A block of mass 0.678 kilograms is suspended from a spring with a spring constant of 28.6 Newtons per meter. When the block is at rest, it is located 1.44 meters above the floor. The block is then set into oscillation by being lifted to a height of 1.83 meters and released. When data is collected, time t = 0 is taken to be when the block is at the lowest point of its motion. Where is the block at 0.69 seconds? Hint: Find the equation of motion.

Amplitude: .39m
Equilibrium Position (x₀): 1.44m
Frequency (f) = 1.034 Hz

Homework Equations

x=A*sin(2*pi*f*t + phi) + x₀

The problem said equations of motion so any of the constant acceleration equations could be useful, but I don't see how they would fit in here.

The Attempt at a Solution

I tried setting phi to equal zero, and that answer came out to be wrong. I don't really know where to start, I know I need to find phi to get the answer, but I'm not sure how to find it.

Note: I don't want the answer, I just need some guidance to getting started. So please don't just post the answer up for me, that won't help on the test, I want to know how to do it, not just the right answer, so no answers, just a place to start would be greatly appreciated, thanks!

 

[BlackHayze]

UPDATE: I used the bottom position to find phi doing:

1.05 = .39m*sin(0+phi)+1.44
(2*pi*f*t = 0 because t=0 at the lowest position)
so -.39 = .39*sin(phi)
-1=sin(phi)
sin^-1(.1)=phi
-90=phi

But using -90 for phi I still get the wrong answer, can anyone confirm that that is the right way to get phi?

If it is, where am I going wrong? I just plug phi into the equation listed above for SHM and got 1.17m, which was wrong.

 

[denverdoc]

One small mistake. Don't go outside the SHM formula.

You want the eqn to be in the form of x(t)=sin (wt + ?).
You are given initial conditions to help with the ?

 

[BlackHayze]

I don't recognize the equation you're using. I don't see how angular velocity would come into play here (assuming that the w is omega.)

I stuck with x=A*sin(2*pi*f*t + phi) + equilibrium position (we were never given the equation you used for SHM,) 2*pi*f*t was zero since time was zero, so that equation was still inside the SHM formula.

 

[denverdoc]

Because simple harmonic motion(SHM) is all about angular velocity. The angle part gets confusing when you are talking about side to side oscillation. But because it is a cycle, you can define one sweep as 2*pi radians.

To know how fast an object vibrates, you express this as an angular veocity. In SHM, the speed at which things vibrates is expressed in w.

But here you re told w=0. The object is at rest. From this you can figure out the phase lag: So the more general form of Asin(wt+?) reduces to Asin(?), You figure out ?, you can then figure out all else.You still need to get w from the data given.

 

[BlackHayze]

But this is an object on a spring, so I don't get where the angular velocity comes in? It's not moving in a circle or rotating, it's oscillating up and down.

 

[denverdoc]

Try imagining attaching a pen to the oscillating mass. You run a strip of paper by the pen. The picture drawn will be that of a sine function repeating at a rate which is determined by the stiffness and the mass of the spring.

lets say it vibrates at one cycle per second. f= 1Hz. This is the same as saying it travels through 2 pi radians every second. 2 pi radians/sec is an angular velocity.

That is where the f*2*pi in your equation comes from--it is the angular velocity. At t=0 this term is zero and the phase angle or offset is phi. The amplitude is governed by the initial stratch of the spring, in this case 0.39m.

So we need to solve A(sin(0+phi)=-.39 which implies Sin(0+phi)= -1 Remember the block is at its lowest point.

Sin(-pi/2)=-1 so this is the phase angle (remember we are not using degrees, but radians).

As it turns out the frequency is (sqrt(k/m)/(2 pi). So you can put that into your sin equation and now compute the displacement any time forward.

You can use an offset Xo as you proposed to place it in meters above the ground if that helps keeps thing clear--in this case it would be X(t)=1.44+0.39(sin...)

There are any number of offests and phase angles you could use or one could even used a cosine function instead of a sine function and had no phase angle to worry about.

Hope this helps.

 

[BlackHayze]

Alright, I get what you're saying now, we were saying the same equation, you were just using a different version of it. I understand how to solve the problem now. Thanks!