Simple Harmonic Motion - Determine quadrant for phase angle.

[OmniNewton]

Homework Statement

When completing this problem I am able to find a value for the phase angle but am unsure of how to find the quadrant for the phase angle therefore unable to get the correct phase angle.

Homework Equations

Provided in the question: x = xmcos(wt+ phi)

The Attempt at a Solution

After doing a series of calculations I determined sin(phi) = -0.8

at t =0

v(0) = -xmwsin(w(0) + phi)
v(0) = -vmsin(phi)
4.0cm/s = (-5.0 cm/s)sin(phi)
sin(phi) = -4/5 = -0.80

Therefore, one answer is -0.93 rad for the phase angle, phi. However other phase angles would also make sin(phi) = -0.8 such as -2.21. How do I determine which angle in radians is correct?

 

[TSny]

One way you might decide is to consider the sign of the acceleration at t = 0.

 

[OmniNewton]

One way you might decide is to consider the sign of the acceleration at t = 0.

Yeah I notice that since when I did my calculations I used t =0 that the acceleration at that point was positive do to the slope of the velocity graph being positive. However how would I determine the angle range? I think it may be important to also know the velocity is positive at t = 0 perhaps.

 

[TSny]

To make sure you are on the right track, would you mind showing how you got sin(φ) = -.80? You should not have needed to know anything about the acceleration to get this.

From the graph, is the acceleration positive or negative at t = 0. This should give you some more information about which quadrant φ is in.

 

[OmniNewton]

To make sure you are on the right track, would you mind showing how you got sin(φ) = -.80? You should not have needed to know anything about the acceleration to get this.

From the graph, is the acceleration positive or negative at t = 0. This should give you some more information about which quadrant φ is in.

Thank you for the response here are my calculations:

at t =0

v(0) = -xmwsin(w(0) + phi)
v(0) = -vmsin(phi)
4.0cm/s = (-5.0 cm/s)sin(phi)
sin(phi) = -4/5 = -0.80

EDIT: The acceleration is positive at t = 0 and so is the velocity. How do these quantities apply to the quadrant?

 

[TSny]

v(0) = -xmwsin(w(0) + phi)
v(0) = -vmsin(phi)
4.0cm/s = (-5.0 cm/s)sin(phi)
sin(phi) = -4/5 = -0.80

Great!

The acceleration is positive at t = 0

How do you determine the acceleration from a velocity vs time graph?

 

[OmniNewton]

Great!How do you determine the acceleration from a velocity vs time graph?

acceleration was determined by the slope of the velocity vs time graph. At t = 0 I had observed a negative slope therefore acceleration is actually negative at t=0. Sorry I made an error. How does this negative acceleration relate to phase angle?

 

[TSny]

What is the mathematical expression for the acceleration as a function of time? What does it reduce to at t = 0?

 

[OmniNewton]

What is the mathematical expression for the acceleration as a function of time? What does it reduce to at t = 0?

a(t) = -xmw^2 cos(wt + phi)
a(0) = -xm(w^2)cos(phi)
a(0) = -xw^2

 

[TSny]

a(t) = -xmw^2 cos(wt + phi)
a(0) = -xm(w^2)cos(phi)

OK to here.

a(0) = -xw^2

What happened to cos(phi)? [EDIT: Oh, I see. You dropped the subscript m on the position. OK]

 

[OmniNewton]

OK to here.

What happened to cos(phi)?

Well I substituted the displacement function x(0)=xmcos(phi) into the velocity equation. I just called this equation x instead of x(0)

 

[TSny]

OK. From a(0) = -xm(w^2)cos(phi) can you deduce anything about the sign of cos(phi)?

 

[OmniNewton]

OK. From a(0) = -xm(w^2)cos(phi) can you deduce anything about the sign of cos(phi)?

since xm and w^2 are always positive the cos(phi) is also positive. Therefore, according to the CAST rule since cos(phi) is positive the quadrant must be either 1 or 4 but sin(phi) is negative so therefore phi is in quadrant for so phi= -0.92 radians (to 2 significant figures) is the answer?

 

[TSny]

Yes, 4th quadrant. -0.92 rad looks correct.

 

[OmniNewton]

Yes, 4th quadrant. -0.92 rad looks correct.

Actually did I make an error, should cos(phi) actually be negative do to the negative in front so the answer should be quadrant 3? since sin and cos are both negative in quadrant 3?

 

[TSny]

Sorry, I'm not following how you are getting that cos(phi) should be negative. What is the sign of a(0)?

 

[OmniNewton]

Sorry, I'm not following how you are getting that cos(phi) should be negative. What is the sign of a(0)?

Not a problem let me try to explain.

xm is a positive quantity. Angular frequency, w , a positive quantity
a(0) = -xm(w^2)cos(phi)
a(0) = -(positive)(positive)^2(cos(phi))
negative * positive * positive = negative
that is,
a(0) = -cos(phi)

 

[TSny]

Not a problem let me try to explain.

xm is a positive quantity. Angular frequency, w , a positive quantity
a(0) = -xm(w^2)cos(phi)
a(0) = -(positive)(positive)^2(cos(phi))
negative * positive * positive = negative
that is,
a(0) = -cos(phi)

Exactly.
So, what is the sign of cos(phi)?

 

[OmniNewton]

Exactly.
So, what is the sign of cos(phi)?

Is it not negative? Or am I perhaps mislead by the negative sign in front of cos(phi)?

 

[TSny]

If cos(phi) were negative, what would be the sign of -cos(phi)?

 

[OmniNewton]

Is it not negative?

If cos(phi) were negative, what would be the sign of -cos(phi)?

Oh I see where you are coming from. My mistake, from the graph it was determined at t = 0 the acceleration is negative in order for the acceleration to remain negative sin(phi) must be positive therefore phi is in quadrant 4.

 

[TSny]

Oh I see where you are coming from. My mistake, from the graph it was determined at t = 0 the acceleration is negative in order for the acceleration to remain negative sin(phi) must be positive therefore phi is in quadrant 4.

Yes, if you meant to say cos(phi) instead of sin(phi).

 

[OmniNewton]

Yes, if you meant to say cos(phi) instead of sin(phi).

Yes that is correct, cos(phi) instead of sin(phi).

Thank you so much for your help now I understand how I can always determine the quadrant in such questions.
OmniNewton

 

[TSny]

OK. Good work.