1. ## need help here

Q1) An oscillating signal voltage is given by E=125cos(wt-&)millivolts, where the angular frequency is w=123pi,phase angle is &=pi/2,and t= time in seconds. The triggering mechanism of an oscilloscope starts the sweep when E= 60mV. what is the smallest postive value of t for which the triggering occurs?

Q2)The displacement d of a piston is given by
d=sinwt+1/2sin2wt

2. Originally Posted by cheesepie
Q1) An oscillating signal voltage is given by E=125cos(wt-&)millivolts, where the angular frequency is w=123pi,phase angle is &=pi/2,and t= time in seconds. The triggering mechanism of an oscilloscope starts the sweep when E= 60mV. what is the smallest postive value of t for which the triggering occurs?
$
E(t)=125\cos(123 \pi t-\pi/2)\ \mbox{mV}
$

The trigger event occurs when $t$ is the smallest positive root of:

$
125\cos(123 \pi t-\pi/2)=60
$

Now:

$
cos(\theta-\pi/2)=sin(\theta)
$
,

so we are interested in the roots of:

$
125\sin(123 \pi t)=60
$

or:

$
\sin(123 \pi t)=60/125=0.48
$

The smallest positive root is what is usual give by the arcsin function on

$
123 \pi t=\arcsin(0.48)\approx 0.5
$
,

so:

$
t\approx0.5/(123\pi) \approx 0.00129 \mbox{seconds}
$

RonL

3. Originally Posted by cheesepie

Q2)The displacement d of a piston is given by
d=sinwt+1/2sin2wt
Yes but what is the question?

RonL

Q2)The displacement d of a piston is given by
d=sinwt+1/2sin2wt

for what primary solutions of wt less than 2pi is d=o ?

5. ## upz

upz upz upz

6. Originally Posted by cheesepie

Q2)The displacement d of a piston is given by
d=sinwt+1/2sin2wt

for what primary solutions of wt less than 2pi is d=o ?
Since both the terms on the RHS are 0 at wt=0 that is a solution,
Is it the one you want though?

RonL

7. the answer is 0 , pi

but can u show me the step ? thank alot ..

8. Originally Posted by cheesepie
the answer is 0 , pi

but can u show me the step ? thank alot ..
$
d(wt)=\sin(wt)+1/2 \sin(2wt)=0
$

Expand the second sin function using the double angle formula:

$
\sin(wt)+1/2 \sin(2wt)=\sin(wt)+\sin(wt)\cos(wt)$
$=\sin(wt)(1+cos(wt))=0
$

Now the only way that $\sin(wt)(1+cos(wt))$ can be zero is if either
$\sin(wt)=0$ or $(1+cos(wt))=0$, the first of these gives $wt=0 \mbox{ and }\pi$,
and the second gives $wt=\pi$, so the solutions in $[0,2\pi]$ are $wt=0 \mbox{ and }\pi$.

RonL

9. hi ...can u explain this step...not too sure ..thk..

$
\sin(wt)+1/2 \sin(2wt)=\sin(wt)+\sin(wt)\cos(wt)$
$=\sin(wt)(1+cos(wt))=0
$

10. upz upz upz

11. Originally Posted by cheesepie
hi ...can u explain this step...not too sure ..thk..
$
\sin(wt)+1/2 \sin(2wt)=\sin(wt)+\sin(wt)\cos(wt)$
$=\sin(wt)(1+cos(wt))=0
$
Hello,

as Capt.Black has suggested you should use the double angle formula of the sine function:

$\sin(2x)=2 \cdot \sin(x) \cdot \cos(x)$. Therefore:

$\frac{1}{2} \cdot \sin(2x)=\sin(x) \cdot \cos(x)$.

So you can sustitute $\frac{1}{2} \cdot \sin(2x)$ by $\sin(x) \cdot \cos(x)$.

Hope that this is of some help.

Greetings

EB
.

12. thanks for ur reply ..

13. Originally Posted by cheesepie
Hi..thanks for ur reply ..i understand this step
only dont understand sin(wt)+sin(wt)cos(wt)=sin(wt)(1+cos(wt)=0
Hello,

you can factorize this sum: Put the common factor (here it is sin(wt)) before the bracket. You get the contents in paranthese by dividing both summands by sin(wt). So sin(wt)/sin(wt) = 1 (not zero!!!), you'll get the sum in the bracket.

Greetings

EB

14. Got it ..thanks ..