Manipulating Simple Harmonic Motion Equation

[Glorzifen]

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 + $$\varphi$$)
v(t) = -wAsin(wt + $$\varphi$$)

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$$\pi$$f...then w will be halved as well (right?):

f = 1/2T --> .5f --> 2$$\pi$$.5f = .5w

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

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.

 

[Hellabyte]

You are correct in your logic that $$\omega$$ is halved as well. Let's look at what the question is asking. What is the *Maxiumum* velocity? You have written the equation for velocity:

$$v(t) = A_0 \omega Sin(\omega t)$$

You should know that $$A_0$$ is the maximum amplitude.

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


$$v_{max} (t) = A_0 \omega$$

Now what will happen when the amplitude and period are both doubled?

 

[Glorzifen]

Ah. Okay...I didn't know how to deal with the whole sine thing (obviously). So based on this V_max equation the max velocity will remain unchanged. Correct?

 

[Hellabyte]

Exactly. When dealing with maximum values, your sin and cosine functions will generally disappear for the reasons stated above.