When the waves Y1=asinwt and Y2=acoswt are superimposed, find the resultant amplitude?
The resultant amplitude can be found using the formula:
A = √(A1^2 + A2^2 + 2A1A2cosφ)
where A1 and A2 are the amplitudes of Y1 and Y2 respectively, and φ is the phase difference between them.
In this case, A1 = a and A2 = a, since Y1 = asinwt and Y2 = acoswt have the same amplitude.
To find φ, we need to express Y1 and Y2 in terms of a common reference. Let's use cos:
Y1 = a sin wt = a cos (π/2 - wt)
Y2 = a cos wt
Now we can see that φ = π/2 - wt.
Substituting these values into the formula, we get:
A = √(a^2 + a^2 + 2a*a cos(π/2-wt))
Simplifying,
A = √(2a^2(1-cos(π/2-wt)))
A = √(2a^2(1+sinwt))
A = a√(2(1+sinwt))
Therefore, the resultant amplitude is a multiplied by the square root of 2 times (1+sinwt).
To find the resultant amplitude when two wave functions, Y1 = asin(wt) and Y2 = acos(wt), are superimposed, we can use the concept of phasors.
1. Recall that the general equation for a wave can be written as Y = A sin(wt + φ), where A represents the amplitude, w is the angular frequency, t is the time, and φ is the phase angle.
2. We can rewrite Y1 and Y2 using the general wave equation:
Y1 = asin(wt + 0°)
Y2 = acos(wt + 90°)
Note: We express angles in degrees for clarity, but they need to be converted to radians for calculations.
3. Now, let's convert Y1 and Y2 to phasor form:
Y1 = A1 * sin(wt + φ1)
Y2 = A2 * sin(wt + φ2)
Where A1 = a and φ1 = 0° for Y1,
And A2 = a and φ2 = 90° for Y2.
4. To find the resultant amplitude, we'll add the two phasors in vector form.
R = A1 * exp(iφ1) + A2 * exp(iφ2)
R = a * exp(i0°) + a * exp(i90°)
5. Recall Euler's formula, which states that exp(ix) = cos(x) + i * sin(x), where i represents the imaginary unit.
Applying Euler's formula to our equation:
R = a * (cos(0°) + i * sin(0°)) + a * (cos(90°) + i * sin(90°))
R = a * (1 + i * 0) + a * (0 + i * 1)
R = a + a * i
6. Finally, we can find the magnitude (resultant amplitude) of the phasor R:
|R| = √(Re(R)^2 + Im(R)^2)
|R| = √(a^2 + (a * 1)^2)
|R| = √(a^2 + a^2)
|R| = √(2a^2)
|R| = √2 * a
Therefore, the resultant amplitude when Y1 = asin(wt) and Y2 = acos(wt) are superimposed is √2 * a.