Graph of simple harmonic motion

[mizzy]

Homework Statement

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

Homework Equations

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?

Can someone help me please?

THANKS

 

[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.

 

[mizzy]

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.

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

 

[mizzy]

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

is that right?

 

[mizzy]

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

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

 

[xcvxcvvc]

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

is that right?

no. the time at which a phase of pi/2 happens, t, will have T in it.

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

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

no. wt must equal 0 or pi for |a(t)| to be maximum.