GREPhysics.NET: GR8677 #77

GR8677 #77

Problem

GREPhysics.NET Official Solution

Wave Phenomena $\Rightarrow$ }Wave Equation

The problem gives the equation of motion \par $m\ddot{x}=-kx \Rightarrow \ddot{x}=-\frac{k}{m}x=-\omega^2 x$ , where $\omega^2=\frac{k}{m}$ .

The general equation for a wave propagating in time and oscillating in the x direction is
$x(t)=A \sin(\omega t + \phi)$ . ( $A$ is the amplitude, $\omega$ is the angular frequency and $\phi$ is some phase constant.)

This is also the general solution to the differential equation posed above.

Plug in the condition (given by the problem) that \par $x(t)=A/2=A\sin(\omega t +\phi)$ to get $1/2=\sin(\omega t + \phi)$ . Recalling the unit circle, the angle \par $\omega t + \phi = \pi/6$ .

Plug in the argument into the velocity \par
$\dot{x}=\omega A \cos(\omega t+\phi)=\omega A \cos(\pi/6)=\omega A \sqrt{3}/2$ . Recall that $\omega = 2\pi f$ , and thus $\dot{x}=\pi f A\sqrt{3}$ , as in choice (B).

Alternate Solutions

lattes
2008-08-07 13:08:18

Another to solve is: consider Let $x(t)=A\cos{\omega t}$ and $v(t)=-A\omega\sin{\omega t}$ . Squaring both equation and summing them we get: ${(\frac{x}{A})}^{2}+{(\frac{v}{A\omega})}^{2}=1$ . Now let $x=A/2$ and solve for $v$ . Thus $v=\sqrt{3}\pi fA$ . (you should remember that $\omega=2\pi f$ !). A little bit longer solution, but it works.

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rreyes
2005-10-31 08:06:24

an alternative solution is conservation of energy

1/2kA^2=1/2kx^2+1/2mv^2

plugging in x=A/2 gives

v= \sqrt{3k/4m}A = \sqrt{3}\pi fA

yosun
2005-11-01 02:14:55

fyi, add in dollar-sign wrappers around the equations in your comments... and the system processes your latex-compatible syntax into nifty-looking equations.

quote rreyes:

an alternative solution is conservation of energy

$1/2kA^2=1/2kx^2+1/2mv^2$

plugging in $x=A/2$ gives

$v= \sqrt{3k/4m}A = \sqrt{3}\pi fA$

nitin
2006-11-16 13:32:49

Thumbs up mate!

Reply to this comment

Comments

lattes
2008-08-07 13:08:18

lattes
2008-08-07 13:09:33

"Another way to solve the problem is" ....

Reply to this comment

rreyes
2005-10-31 08:06:24

an alternative solution is conservation of energy

1/2kA^2=1/2kx^2+1/2mv^2

plugging in x=A/2 gives

v= \sqrt{3k/4m}A = \sqrt{3}\pi fA

yosun
2005-11-01 02:14:55

nitin
2006-11-16 13:32:49

Thumbs up mate!

Reply to this comment

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LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

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