The r.m.s value of Potential due to superposition of given two alternating potentials `E_(1) = E_(0) sin omega t` and `E_(2) = E_(0) cos omega t` will be

The r.m.s. value of I = I_(1) sin omega t + I_(2) cos omega t is

The r.m.s. value of I = I_(1) sin omega t + I_(2) cos omega t is

Two alternating current are given by l_(1)=l_(0)" sin "omega t and l l_(2)=l_(0) cos (omega +phi) The ratio of rms value is

The amplitude of the vibrating particle due to superposition of two SHMs , y_(1)=sin (omega t+(pi)/(3)) and y_(2)=sin omega t is :

The resultant amplitude due to superposition of three simple harmonic motions x_(1) = 3sin omega t , x_(2) = 5sin (omega t + 37^(@)) and x_(3) = - 15cos omega t is

Three light waves combine at a certain point where thcir electric field components are E_(1) = E_(0) sin omega t , E_(2)= E_(0) sin(omega t +60^(@)) , E_(3)= E_(0) sin(omega t -30^(@)) . Find their resultant component E(t) at that point.

Two coherent sources emit light waves which superimpose at a point where these can be expressed as E_(1) = E_(0) sin(omega t + pi//4) E_(2) = 2E_(0) sin(omega t - pi//4) Here, E_(1) and E_(2) are the electric field strenghts of the two waves at the given point. If I is the intensity of wave expressed by field strenght E_(1) , find the resultant intensity

Two waves at a point are represented by E_(1)= E_(0) sin omega t and E_(2)=E_(0) sin ( omega + phi) . There will eb destructive interference at this point is

In the given figure, the instantaneous value of alternating e.m.f. is e = 14.14 sin omega t . The reading of voltmeter in volt will be