# Manipulating Simple Harmonic Motion Equation

## 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
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.

Related Introductory Physics Homework Help News on Phys.org
You are correct in your logic that $$\omega$$ is halved as well. Lets 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?

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?

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