A 342-g mass is attach to a spring and undergoes SHM.

In summary: The spring constant is supposed to be k=m/A, but you get k=.165m. This might be an error in the textbook or on your part. Confirm that the given values are correct.
  • #1
utorontoconnor
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Homework Statement


A 342-g mass is attach to a spring and undergoes SHM.
Its max accelerations is 18.6m/s^2
Its max velocity is 1.75m/s

Homework Equations


x(t)=Acos(wt)
x'(t)=-wAsin(wt)
x''(t)=-w^2Acos(wt)
w=sqrt(k/m)

The Attempt at a Solution


I essentially derived these equations and have no idea where to go or how to equate them... Using Wolfson "University Physics" 3rd edition. [/B]
 
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  • #2
What is the question?
 
  • #3
Oops sorry... sequentially determine A) the angular frequency B) the amplitude, and C) the spring constant
 
  • #4
For simple harmonic motion the general solution is ##x(t)=Acos(\omega t+\phi)## where ##\omega=\sqrt{k/m}## and ##\phi## is a phase. I don't know why you are assuming that ##\phi=0##. Please, also give the problem statement. What is the question?
 
  • #5
Terribly sorry... I'm not sure what my first steps would be to find the angular frequency, ω, given a mass of 342-g attached to a spring undergoing simple harmonic motion, a max acceleration of 18.6m/s^2, and a maximum speed* of 1.75 m/s. Essentially, I don't know how to use F=-kx, x(t)=Acos(ωt + φ) and the derivatives to solve the aforementioned, angular frequency.
 
  • #6
Since you're given the maximum values for the acceleration and velocity and you weren't given a particular starting position, you don't need to worry about a phase angle. You can assume that the phase angle is zero. Your set of relevant equations is fine.

From your relevant equations, pick out the expressions that represent the maximum acceleration and velocity.
 
  • #7
x'(t)=-wAsin(wt)
So, if I want to maximize my velocity, I'd need to make sin(wt)=1, since sine oscillates between -1, and 1, the appropriate time would be at π/2? Hence, I would be left with x'(t)max=-wA(1)
then, x''(t)=-w^2Acos(wt) would need to have cos(wt)=1 as well in order to get max velocity, because it's acceleration is greatest when x is further away form equilibrium?
If so, then x''(t)=-w^2A

which then leads me to where...? Sorry I'm have a tough go at wrapping my head around this. I'm sure its simpler than I'm making it out to be.
 
  • #8
utorontoconnor said:
x'(t)=-wAsin(wt)
So, if I want to maximize my velocity, I'd need to make sin(wt)=1, since sine oscillates between -1, and 1, the appropriate time would be at π/2? Hence, I would be left with x'(t)max=-wA(1)
then, x''(t)=-w^2Acos(wt) would need to have cos(wt)=1 as well in order to get max velocity, because it's acceleration is greatest when x is further away form equilibrium?
If so, then x''(t)=-w^2A

which then leads me to where...? Sorry I'm have a tough go at wrapping my head around this. I'm sure its simpler than I'm making it out to be.
Yes, the coefficient of the sine or cosine sets the maximum amplitude of the term since the maximum value for sine or cosine is 1.

You've been given values for the maximum acceleration and velocity. Given them variable names so you don't have to drag around a lot of digits. Maybe call them Amax and Vmax. Write out the expressions for Amax and Vmax that you've extracted from your relevant equations.
 
  • #9
Vmax= -ωA
Amax=-ω^2A
can I then equate ω=Vman/A? and the sub that into my Amax ω?
 
  • #10
Or am i able to isolate for A and then replug that into either equation to solve for ω?
 
  • #11
utorontoconnor said:
Vmax= -ωA
Amax=-ω^2A
can I then equate ω=Vman/A? and the sub that into my Amax ω?
Yes.
utorontoconnor said:
Or am i able to isolate for A and then replug that into either equation to solve for ω?
Either way, it's up to you. The important thing is that you have a path forward towards your goal.

Note that you can drop the negative signs from the terms since they represent the maximum values, hence are absolute values (and you could have chosen sine and cosine to be -1 rather than +1 to make those expressions positive).

You might want to try out the x2 and x2 icons in the edit window for producing subscripts
 
  • #12
Thank you! Really appreciate the guidance... Although, I appear to be off by a factor 1/10:
Vmax=wA
Amax=w^2A
where w= Vmax/A, thus Amax=[(Vmax/A)^2]xA
Hence, A= Vmax^2/Amax= 3.0625/18.6= .165m.
Hence, w=1.75/.165= 10.60... though my text says it should be 1.06 or 10.6^-1...
Where would my error be for that?
 
  • #13
I agree with your result. Confirm that the given values are correct.

Textbooks have been known to have typos and errors, particularly if problems are "updated" for new editions and the answer key is not updated, too. For example, if Amax happened to be 1.86 instead of 18.6 m/s2 then that could account for the difference.

See how the value you get for the spring constant compares with what they say it should be.
 
  • #14
Sorry- I think the textbook just has some typos...
 
  • #15
Sorry, that was a waste of your time! Thanks again for all of your help! I look forward to your help again, hopefully!
Cheers
 
  • #16
No worries. We're here to help :smile:
 

FAQ: A 342-g mass is attach to a spring and undergoes SHM.

1. What is SHM?

SHM stands for Simple Harmonic Motion. It is a type of periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position and acts in the opposite direction of the displacement.

2. How is SHM different from other types of motion?

SHM is different from other types of motion because it follows a sinusoidal pattern and has a constant period and amplitude. This means that the motion repeats itself at regular intervals and the maximum displacement from equilibrium remains constant.

3. What factors affect the period of SHM?

The period of SHM is affected by two main factors: the mass attached to the spring and the stiffness of the spring. A larger mass or a stiffer spring will result in a longer period, while a smaller mass or a less stiff spring will result in a shorter period.

4. How is the amplitude of SHM determined?

The amplitude of SHM is determined by the initial displacement of the mass from the equilibrium position. The greater the initial displacement, the larger the amplitude of the motion will be.

5. Can SHM ever be perfectly achieved in real life?

In theory, SHM can be perfectly achieved as long as there is no external influence or damping force acting on the system. However, in reality, it is difficult to achieve perfect SHM due to factors such as friction and air resistance, which can affect the motion of the mass-spring system.

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