Write expressions for simple harmonic motion

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SUMMARY

The discussion focuses on writing expressions for simple harmonic motion (SHM) with specific parameters. For part (a), the correct expression is confirmed as x(t) = (10 cm)cos(10πt), where the amplitude is 10 cm and frequency is 5.0 Hz, indicating a phase angle of zero. In part (b), the expression x(t) = (2.5 cm)cos(5.0t + 3π/2) is derived, with an amplitude of 2.5 cm and angular frequency of 5.0 1/s, showing a phase shift that transforms the cosine function into a sine function. The solutions align with the provided answer book, confirming their accuracy.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM) concepts
  • Familiarity with trigonometric functions and their properties
  • Knowledge of angular frequency and its calculation
  • Ability to manipulate phase angles in SHM equations
NEXT STEPS
  • Study the derivation of SHM equations in detail
  • Learn about the relationship between frequency and angular frequency
  • Explore the effects of phase shifts on SHM graphs
  • Practice solving SHM problems with varying parameters and conditions
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Students studying physics, particularly those focusing on mechanics and wave motion, as well as educators looking to reinforce concepts of simple 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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