Manipulating Simple Harmonic Motion Equation

  • Thread starter Glorzifen
  • Start date
  • #1
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Homework Statement


An object moves with simple harmonic motion. If the amplitude and the period are both doubled, the object's maximum speed is:
A) Quartered
B) Halved
C) Quadrupled
D) Doubled
E) Unchanged


Homework Equations


x(t) = Acos(wt + [tex]\varphi[/tex])
v(t) = -wAsin(wt + [tex]\varphi[/tex])


The Attempt at a Solution


Since 2 double the period (T) is to halve the frequency (f) because of (f = 1/T), and since omega (w) = 2[tex]\pi[/tex]f...then w will be halved as well (right?):

f = 1/2T --> .5f --> 2[tex]\pi[/tex].5f = .5w

So that gives me: v(t) = -.5wAsin(.5w + [tex]\varphi[/tex])
And with the amplitude doubled: v(t) = -wAsin(.5w + [tex]\varphi[/tex])

Not sure if I did any of that correctly and I'm not sure what that means for my speed...is it halved? Any help would be appreciated. Thanks.
 

Answers and Replies

  • #2
53
0
You are correct in your logic that [tex]\omega[/tex] is halved as well. Lets look at what the question is asking. What is the *Maxiumum* velocity? You have written the equation for velocity:

[tex] v(t) = A_0 \omega Sin(\omega t) [/tex]

You should know that [tex] A_0 [/tex] is the maximum amplitude.

Because the maximum value of any Sin function is 1, that means that the maximum velocity will be given by


[tex] v_{max} (t) = A_0 \omega [/tex]

Now what will happen when the amplitude and period are both doubled?
 
  • #3
25
0
Ah. Okay...I didn't know how to deal with the whole sine thing (obviously). So based on this Vmax equation the max velocity will remain unchanged. Correct?
 
  • #4
53
0
Exactly. When dealing with maximum values, your sin and cosine functions will generally disappear for the reasons stated above.
 

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