आवर्ती विस्थापन `y = A cos omega t +A cos (omegat +(2pi)/3)+A cos (omegat +(4pi)/3)` का परिणाम होगा -

The resulting amplitude A' and the vebrations S = A cos (omega t) + (A)/(2) cos (omega t + (pi)/(2)) xx (A)/(4) cos (omega t + pi) = (A)/(8) cos (omega t+ (3pi)/(2)) = A' cos (omega t + delta) are…and…respectively.

Find the resulting amplitude A' and the phase of the vibrations delta S = Acos(omegat) + A/2 cos (omegat + pi/2) + A/2 cos (omega t + pi) + A/8 cos (omegat + (3pi)/(2)) = A' cos (omega t + delta)

Find the resulting amplitude A' and the phase of the vibrations delta S = Acos(omegat) + A/2 cos (omegat + pi/2) + A/2 cos (omega t + pi) + A/8 cos (omegat + (3pi)/(2)) = A' cos (omega t + delta)

If current 1_(1) = 3A sin omegat and 1_(2)= 4A cos omegat, then 1_(3) is

If x=y cos((2pi)/(3))=z cos ((4pi)/(3)) then xy+yz+zx=

For simple harmonic vibrations y_(1)=8cos omegat y_(2)=4 cos (omegat+(pi)/(2)) y_(3)=2cos (omegat+pi) y_(4)=cos(omegat+(3pi)/(2)) are superimposed on one another. The resulting amplitude and phase are respectively

For simple harmonic vibrations y_(1)=8cos omegat y_(2)=4 cos (omegat+(pi)/(2)) y_(3)=2cos (omegat+pi) y_(4)=cos(omegat+(3pi)/(2)) are superimposed on one another. The resulting amplitude and phase are respectively

A vibratory motion is represented by x=24cosomegat+A cos(omegat+(pi)/(2))+A cos(omegat+pi)(A)/(2)cos(omegat+(3pi)/(2)) the resultant amplitude of the motion is