# Graph of simple harmonic motion

1. Aug 22, 2010

### mizzy

1. The problem statement, all variables and given/known data

A particle moves with amplitude A and period T (see figure). Express the following in terms of A and T and numberical constants.
a) The time at which the phase is pi/2
b) maximum v
c) max a
d) time at which acceleration is a maximum

2. Relevant equations

3. The attempt at a solution
I got b and c, but i'm not sure about a and d.

For a, they ask for the time at which phase is pi/2. I know T = 2pi. For d, when acceleration is max, is time 0?

THANKS

#### Attached Files:

• ###### physics.JPG
File size:
12.1 KB
Views:
114
2. Aug 22, 2010

### xcvxcvvc

T does not equal 2pi

$$\omega T = 2 \pi$$
and
$$\omega = \frac{2\pi}{T}$$
where omega is the radian frequency, because when you evaluate the sinusoid at t = T, the argument equals 2pi.

$$x(t) = cos(\omega t)$$
$$x(t) = cos(\frac{2\pi}{T}t)$$

Using this last equation, we can solve both a and d. For a, find a t that makes the argument pi/2. For d, differentiate twice to find acceleration and find the argument that maximizes acceleration.

you can use a more conceptual approach for both, however. For example, if you know how a cosine behaves, you can instantly know A just by looking at the graph.

3. Aug 22, 2010

### mizzy

I'm not getting what you mean by finding a t that makes it pi/2. We don't have any values of t

4. Aug 22, 2010

### mizzy

for a, time has to be 1/4t.

is that right?

5. Aug 22, 2010

### mizzy

differentiating the equation twice gives acceleration = -A omega^2 cos (omega t)

amax = Aomega^2, so t has to be equal to 1?

6. Aug 23, 2010

### xcvxcvvc

no. the time at which a phase of pi/2 happens, t, will have T in it.
no. wt must equal 0 or pi for |a(t)| to be maximum.