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problem of two vibrations(using trignometry)

need some help with this problem folks two vibrations,x1=3sin(10t +pi/6) x2=2cos(10t + pi/6).where t is in seconds and are superimposed.determine the time at which the amplitude of the resultant vibration,x1+x2,first reaches the value of 2 .

solution

3sin(10t+pi/6) + 2cos(10t - pi/6) =2

simplifying 3sin(10t+pi/6) using compound angle formulae
we get
2.598sin10t + 1.5cos10t (A)

now
simplifying 2cos(10t -pi/6) using compound angle formulae
we get
1.732cos10t + sin 10t (B)

adding (A) & (B)

2.598sin10t + 1.5cos10t + 1.732cos10t + sin 10t

therefore 3.598sin10t + 3.232cos10t

the above equation is in the form asinx + bcosx
now we know that asinx + bcosx=Rsin(x + alpha)

R=sq root of 12.9456 +10.4458
R=4.8364
alpha=arctan 3.232/3.58
=0.73186

therefore writing the equation in the form Rsin(x + alpha)

we get

3.598sin10t + 3.232cos10t =4.8364sin(x+0.73186)
substituting 10t=x

from the given problem we know that the sum of two vibrations is =2
therefore
4.8364sin(x+0.73186)=2
sin(x+0.73186)=0.4135
(x+0.73186)=arcsin 0.4135
(x+0.73186)=0.4263
x= - 0.30556
thats the answer i get and i know it is incorrect.
i did not comprehend the idea of adding 2pi to (10t+pi/6).
at which step do i it?

i have solved the equations using compound angle formula and using x=10t i have found the value of "t".but the value of "t" i get is negative.

any help regarding the above will be appreciated.

Reply 1

AlphaNumeric

sin and cos are periodic, so whatever value of

10t + \frac{\pi}{6}

you're using, add

2\pi

to it, and then work out t. If it's still negative, add another

2\pi

, and keep doing so till you get a positive value of t.

Reply 2

mninjas

AlphaNumeric

sin and cos are periodic, so whatever value of

10t + \frac{\pi}{6}

you're using, add

2\pi

to it, and then work out t. If it's still negative, add another

2\pi

, and keep doing so till you get a positive value of t.

i do not know where to introduce yur idea,so can you please help me out.

Reply 3

AlphaNumeric

You know that sin(x+2pi) = sin(x) and cos(x+2pi) = cos(x).

Therefore if you add or subtract 2pi, 4pi, 6pi etc from any of your solutions for x you'll end up with a sin and cos value which is the same. For instance

sin(x+0.73186)=0.4135 gives you x= - 0.30556
but also
sin(x+0.73186 -2pi) = 0.4135, so you get x= - 0.30556 + 2pi, which is going to be bigger than 0.