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

1. ## 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?

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 ......

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:

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

4. 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.

5. 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

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