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.
RedBarchetta
Messages
49
Reaction score
1

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.
 
Physics news on Phys.org
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top