Write expressions for simple harmonic motion

AI Thread Summary
The discussion focuses on writing expressions for simple harmonic motion with specified parameters. For part (a), the expression derived is x(t) = (10 cm)cos(10πt), which aligns with the given amplitude and frequency. In part (b), the derived expression is x(t) = (2.5 cm)cos(5t + 3π/2), indicating a phase shift that transforms the cosine function into a sine function. The correctness of the solutions is confirmed, with part (a) being accurate based on frequency and part (b) being equivalent to the answer in the solution key. The conversation highlights the relationship between phase shifts and the transformation of trigonometric functions in harmonic motion.
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Homework Statement


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.

Homework Equations



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

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...?

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

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

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

So...?

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

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

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

Any help is appreciated, thank you.
 
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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,

<br /> cos(x + \frac{3\pi}{2}) = sin (x)<br />
 
Coto said:
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,

<br /> cos(x + \frac{3\pi}{2}) = sin (x)<br />

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