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

**ssadi** · December 29th 2008, 11:53 AM

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?

**mr fantastic** · December 29th 2008, 12:25 PM

Originally Posted by ssadi

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?

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

You should think about why ......

**hatsoff** · December 29th 2008, 01:06 PM

Originally Posted by ssadi

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?

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:

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

**ssadi** · December 29th 2008, 01:20 PM

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.

**ssadi** · December 29th 2008, 01:23 PM

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

**hatsoff** · December 29th 2008, 01:44 PM

Originally Posted by ssadi

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

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

**mr fantastic** · December 29th 2008, 05:38 PM

Originally Posted by ssadi

Tell me why, I am new at the chapter and is not yet acclimatised.

From the data given in the question, you can assume the centre of motion to be at x = 0 and for the particle to start from this position (ie. x = 0 at t = 0). This avoids the detailed calculations that have been provided ....

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