# Write expressions for simple harmonic motion

1. Nov 11, 2008

### RedBarchetta

1. The problem statement, all variables and given/known data
Write expressions for simple harmonic motion (a) with amplitude 10 cm, frequency 5.0 Hz, and maximum displacement at t=0; and (b) with amplitude 2.5 cm, angular frequency 5.0 1/s, and maximum velocity at t=0.

2. Relevant equations

$$\begin{gathered} x(t) = A\cos (\omega t + \varphi ) \hfill \\ \omega = 2\pi f \hfill \\ f = \frac{1} {T} \hfill \\ \end{gathered}$$

3. The attempt at a solution

(a)

A=10 cm
f=5.0 Hz

Since the amplitude equals the max displacement at a given t(in this instance t=0), this tells us that the phase angle is zero. So our equation should be....?

$$x(t) = (10cm)\cos \left[ {(10\pi s^{ - 1} )t} \right]$$

(b)
A=2.5 cm
w=5.0 s^-1

$$\begin{gathered} V(x) = - A\omega \sin (\omega t + \varphi ) \hfill \\ V(0) = A\omega = V_{\max } \hfill \\ V(0) = - A\omega \sin (\varphi ) \hfill \\ A\omega = - A\omega \sin (\varphi ) \hfill \\ - 1 = \sin (\varphi ) \hfill \\ \varphi = \tfrac{{3\pi }} {2} \hfill \\ \end{gathered}$$

So.....?

$$x(t) = (2.5cm)\cos \left[ {(5.0s^{ - 1} )t + \tfrac{{3\pi }} {2}} \right]$$

Do these look right? Here is what my answer book gives:

$$\begin{gathered} (a):x(t) = (10cm)\cos \left[ {(\pi s^{ - 1} )t} \right] \hfill \\ (b):x(t) = (2.5cm)\sin \left[ {(5s^{ - 1} )t} \right] \hfill \\ \end{gathered}$$

Any help is appreciated, thank you.

2. Nov 11, 2008

### Coto

At a glance, your part (a) should be right considering the frequency given. Your part (b) is equivalent to the answer they state.

A phase shift corresponds to a shift in where the peaks occur. You can consider then, that the phase shift can shift a cosine into a sine or a sine into a cosine. In this sense,

$$cos(x + \frac{3\pi}{2}) = sin (x)$$

3. Nov 11, 2008

### RedBarchetta

Thanks Coto. I had a feeling that the part a solution key was incorrect.