Answer to Question #89118 in Physics for Shivam Nishad

Question

Two collinear harmonic oscillations, having amplitudes 4 and 3 units, respectively, are superposed Calculate the amplitude of the resultant oscillation, their frequencies are equal but they have a phase difference of 90.

Answer 1

$$x_1(t)=4 sin(\omega t)$$ $$x_2(t)=3 sin(\omega t + \pi / 2) = 3 cos(\omega t)$$ $$x(t)=x_1(t)+x_2(t)=4sin(\omega t)+3cos(\omega t)=A \sdot sin(\omega t +\phi)$$

where

$$A=\sqrt{3^2+4^2}=5$$ $$\phi=arcsin(3/A)=arccos(4/A)$$

so, the resultant oscillation:

$$x(t)=5 sin(\omega t + arcsin(3/5))$$

second option:

$$x_2​(t)=3sin(ωt-π/2)=-3cos(ωt)$$ $$A=\sqrt{(-3)^2+4^2}=5$$ $$\phi=arcsin(-3/A)=-arcsin(3/A)$$

then, the resultant oscillation:

$$x(t)=5 sin(\omega t - arcsin(3/5))$$