# Mechanics stuck on a sum on simple harmonic motion:help:

• Dec 29th 2008, 11:53 AM
Mechanics stuck on a sum on simple harmonic motion:help:
A particle P is moving on a straight line with S.H.M. of period pi/3 s. Its maximum speed is 5 m/s. Calculate the amplitude of the motion and the speed of P 0.2s after passing through the centre of oscillation.
My workings:
T=2pi/w So w=6
x=0, v=5
5^2=6^2(a^2-0) so a=5/6 (matched with book)
t=0.2s
v=-awsinwt=-5/6*6sin (1.2)=4.66
The answer given in book is 1.81.
Somebody help. What did go wrong with this very simple sum?http://static.thestudentroom.co.uk/i...ilies/help.gif
• Dec 29th 2008, 12:25 PM
mr fantastic
Quote:

A particle P is moving on a straight line with S.H.M. of period pi/3 s. Its maximum speed is 5 m/s. Calculate the amplitude of the motion and the speed of P 0.2s after passing through the centre of oscillation.
My workings:
T=2pi/w So w=6
x=0, v=5
5^2=6^2(a^2-0) so a=5/6 (matched with book)
t=0.2s
v=-awsinwt=-5/6*6sin (1.2)=4.66
The answer given in book is 1.81.
Somebody help. What did go wrong with this very simple sum?http://static.thestudentroom.co.uk/i...ilies/help.gif

x = (5/6) sin (6t) NOT (5/6) cos (6t).

You should think about why ......
• Dec 29th 2008, 01:06 PM
hatsoff
Quote:

A particle P is moving on a straight line with S.H.M. of period pi/3 s. Its maximum speed is 5 m/s. Calculate the amplitude of the motion and the speed of P 0.2s after passing through the centre of oscillation.
My workings:
T=2pi/w So w=6
x=0, v=5
5^2=6^2(a^2-0) so a=5/6 (matched with book)
t=0.2s
v=-awsinwt=-5/6*6sin (1.2)=4.66
The answer given in book is 1.81.
Somebody help. What did go wrong with this very simple sum?http://static.thestudentroom.co.uk/i...ilies/help.gif

Interesting. I've never heard of this before, but according to wikipedia, simple harmonic motion is given by:

$x(t)=A\cos{(2\pi ft+\phi)}$

where $x(t)$ is displacement, $t$ is time, $A$ is amplitude, $f$ is frequency, and $\phi$ is phase.

We also note that period $T$ is given by:

$T=\frac{1}{f}$

This means that $f=\frac{3}{\pi}$. So:

$x(t)=A\cos{(2\pi \frac{3}{\pi}t+\phi)}=A\cos{(6t+\phi)}$

To find velocity, we take the derivative:

$v(t)=x'(t)=-6A\sin{(6t+\phi)}$

To find velocity extrema, we take the derivative of $v(t)$ and set it equal to zero:

$v'(t)=a(t)=-36A\cos{(6t+\phi)}$

$-36A\cos{(6t+\phi)}=0$

$6t+\phi=\frac{(2n-1)\pi}{2}$

$t=\frac{(2n-1)\pi-2\phi}{12}$

Let's plug that into our velocity function:

$-6A\sin{[6\frac{(2n-1)\pi-2\phi}{12}+\phi]}=5$

$A=-\frac{5}{6\sin{[\frac{(2n-1)\pi}{2}]}}$

$A=\{-\frac{5}{6\sin{[\frac{\pi}{2}]}},-\frac{5}{6\sin{[\frac{3\pi}{2}]}}\}$

$A=\{-\frac{5}{6},\frac{5}{6}\}$

Since the amplitude must be a positive value, we can just say:

$A=\frac{5}{6}$

Now, the center of oscillation is another way of saying $x(t)=0$. So:

$x(t)=\frac{5}{6}\cos{(6t+\phi)}=0$

$\cos{(6t+\phi)}=0$

$6t+\phi=\frac{(2n-1)\pi}{2}$

$t=\frac{(2n-1)\pi-2\phi}{12}$

Let's let $t=0$ and $n=1$:

$0=\frac{([2(1)-1]\pi-2\phi}{12}$

$\phi=\frac{\pi}{2}$

And we plug that into our velocity function:

$v(t)=-5\sin{(6t+\frac{\pi}{2})}$

Then plug in $t$:

$v(0.2)=-5\sin{(6[0.2]+\frac{\pi}{2})}\approx-1.81$

But of course we know that speed is relative, and so the value is actually:

$v(t_0+0.2)\approx\pm1.81\;|\;x(t_0)=0$
• Dec 29th 2008, 01:20 PM
Quote:

Originally Posted by mr fantastic
x = (5/6) sin (6t) NOT (5/6) cos (6t).

You should think about why ......

Tell me why, I am new at the chapter and is not yet acclimatised.
• Dec 29th 2008, 01:23 PM
Quote:

Originally Posted by hatsoff
Interesting. I've never heard of this before, but according to wikipedia, simple harmonic motion is given by:

$x(t)=A\cos{(2\pi ft+\phi)}$

where $x(t)$ is displacement, $t$ is time, $A$ is amplitude, $f$ is frequency, and $\phi$ is phase.

We also note that period $T$ is given by:

$T=\frac{1}{f}$

This means that $f=\frac{3}{\pi}$. So:

$x(t)=A\cos{(2\pi \frac{3}{\pi}t+\phi)}=A\cos{(6t+\phi)}$

To find velocity, we take the derivative:

$v(t)=x'(t)=-6A\sin{(6t+\phi)}$

To find velocity extrema, we take the derivative of $v(t)$ and set it equal to zero:

$v'(t)=a(t)=-36A\cos{(6t+\phi)}$

$-36A\cos{(6t+\phi)}=0$

$6t+\phi=\frac{(2n-1)\pi}{2}$

$t=\frac{(2n-1)\pi-2\phi}{12}$

Let's plug that into our velocity function:

$-6A\sin{[6\frac{(2n-1)\pi-2\phi}{12}+\phi]}=5$

$A=-\frac{5}{6\sin{[\frac{(2n-1)\pi}{2}]}}$

$A=\{-\frac{5}{6\sin{[\frac{\pi}{2}]}},-\frac{5}{6\sin{[\frac{3\pi}{2}]}}\}$

$A=\{-\frac{5}{6},\frac{5}{6}\}$

Since the amplitude must be a positive value, we can just say:

$A=\frac{5}{6}$

Now, the center of oscillation is another way of saying $x(t)=0$. So:

$x(t)=\frac{5}{6}\cos{(6t+\phi)}=0$

$\cos{(6t+\phi)}=0$

$6t+\phi=\frac{(2n-1)\pi}{2}$

$t=\frac{(2n-1)\pi-2\phi}{12}$

Let's let $t=0$ and $n=1$:

$0=\frac{([2(1)-1]\pi-2\phi}{12}$

$\phi=\frac{\pi}{2}$

And we plug that into our velocity function:

$v(t)=-5\sin{(6t+\frac{\pi}{2})}$

Then plug in $t$:

$v(0.2)=-5\sin{(6[0.2]+\frac{\pi}{2})}\approx-1.81$

But of course we know that speed is relative, and so the value is actually:

$v(t_0+0.2)\approx\pm1.81\;|\;x(t_0)=0$

I used the formulas from book:
x=acoswt
v=-awsinwt
a=-aw^2coswt=-w^2coswt=-wx^2
Believe me, the sum isn't supposed to be that long:D
• Dec 29th 2008, 01:44 PM
hatsoff
Quote:

I used the formulas from book:
x=acoswt
v=-awsinwt
a=-aw^2coswt=-w^2coswt=-wx^2
Believe me, the sum isn't supposed to be that long:D

Your book's function is just a simplification where the phase is zero. The important thing to remember is that you're not looking for $v(0.2)$. Rather, you're looking for $v(t_0+0.2)\;|\;x(t_0)=0$.
• Dec 29th 2008, 05:38 PM
mr fantastic
Quote: