# Thread: Differentiating area for physics...

1. ## Differentiating area for physics...

Hey guys, im Chris and im new to the forum. Im just getting into the bulk of my ME degree in my sophmore year and things are proving quite difficult.
Anyways, im stuck on the calc 2 problem related to simple harmonic motion.

s(t) = √2E /√K sin ((√K/√M)t + φ)

Find the area trapped between the graph of the function s(t) and the t-axis over one half of the period (one arc of the wave)
the wave between -Q(√M/√K) and (√M/√K)(π – Q)

Heres where im at:

(√2E /√K)(- √M/√K) cos (π – Q) - (√2E /√K)(- √M/√K) cos (– Q)

does it look right? do i need to go further somehow?

2. Originally Posted by chrsr345
Hey guys, im Chris and im new to the forum. Im just getting into the bulk of my ME degree in my sophmore year and things are proving quite difficult.
Anyways, im stuck on the calc 2 problem related to simple harmonic motion.

s(t) = √2E /√K sin ((√K/√M)t + φ)

Find the area trapped between the graph of the function s(t) and the t-axis over one half of the period (one arc of the wave)
the wave between -Q(√M/√K) and (√M/√K)(π – Q)

Heres where im at:

(√2E /√K)(- √M/√K) cos (π – Q) - (√2E /√K)(- √M/√K) cos (– Q)

does it look right? do i need to go further somehow?

You need to find
$\displaystyle \int_{-\phi \sqrt{\frac{M}{K}}}^{(\pi - \phi)\sqrt{\frac{M}{K}}} \sqrt{\frac{2E}{K}} sin \left ( \sqrt{\frac{K}{M}}t + \phi \right ) dt$

Note the limits of integration carefully. You found them correctly, but you didn't apply them into the final answer correctly. For example:
$\displaystyle \sqrt{\frac{K}{M}} \left ( (\pi - \phi)\sqrt{\frac{M}{K}} \right ) + \phi = \pi$
Recalculate your other limit as well.

Also, you can simplify the overall coefficient to
$\displaystyle -\frac{\sqrt{2ME}}{K}$

-Dan

3. ah thanks about the coefficient. But im confused about the $\displaystyle \sqrt{\frac{K}{M}} \left ( (\pi - \phi)\sqrt{\frac{M}{K}} \right ) + \phi = \pi$ How and why did you solve for pi?

that looks just like what i had for my F(a)- F(b)...

(coeff)cos$\displaystyle \sqrt{\frac{K}{M}} \left ( (\pi - \phi)\sqrt{\frac{M}{K}} \right ) + \phi$ - (coeff)cos $\displaystyle \sqrt{\frac{K}{M}} \left (\phi)\sqrt{\frac{M}{K}} \right ) + \phi$ =
(√2E /√K)(- √M/√K) cos (π – Q) - (√2E /√K)(- √M/√K) cos (– Q)

Did i miss something?

4. Originally Posted by chrsr345
ah thanks about the coefficient. But im confused about the $\displaystyle \sqrt{\frac{K}{M}} \left ( (\pi - \phi)\sqrt{\frac{M}{K}} \right ) + \phi = \pi$ How and why did you solve for pi?

that looks just like what i had for my F(a)- F(b)...

(coeff)cos$\displaystyle \sqrt{\frac{K}{M}} \left ( (\pi - \phi)\sqrt{\frac{M}{K}} \right ) + \phi$ - (coeff)cos $\displaystyle \sqrt{\frac{K}{M}} \left (\phi)\sqrt{\frac{M}{K}} \right ) + \phi$ =
(√2E /√K)(- √M/√K) cos (π – Q) - (√2E /√K)(- √M/√K) cos (– Q)

Did i miss something?
I'm saying you put your limits into your F(a) - F(b) incorrectly.
$\displaystyle \sqrt{\frac{K}{M}} \left ( (\pi - \phi)\sqrt{\frac{M}{K}} \right ) + \phi = \sqrt{\frac{K}{M}} \cdot \pi \cdot \sqrt{\frac{M}{K}} - \sqrt{\frac{K}{M}} \cdot \phi \cdot \sqrt{\frac{M}{K}} + \phi = \pi - \phi + \phi = \pi$

You were saying
$\displaystyle \sqrt{\frac{K}{M}} \left ( (\pi - \phi)\sqrt{\frac{M}{K}} \right ) + \phi = \pi - \phi$

Similarly
$\displaystyle \sqrt{\frac{K}{M}} \left ( -\phi \sqrt{\frac{M}{K}} \right ) + \phi = 0$
not
$\displaystyle \sqrt{\frac{K}{M}} \left ( -\phi \sqrt{\frac{M}{K}} \right ) + \phi = -\phi$

-Dan