Answer to Question #89118 in Physics for Shivam Nishad

Question #89118
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.
1
Expert's answer
2019-05-14T09:47:28-0400
x1(t)=4sin(ωt)x_1(t)=4 sin(\omega t)x2(t)=3sin(ωt+π/2)=3cos(ωt)x_2(t)=3 sin(\omega t + \pi / 2) = 3 cos(\omega t)x(t)=x1(t)+x2(t)=4sin(ωt)+3cos(ωt)=Asin(ωt+ϕ)x(t)=x_1(t)+x_2(t)=4sin(\omega t)+3cos(\omega t)=A \sdot sin(\omega t +\phi)

where

A=32+42=5A=\sqrt{3^2+4^2}=5ϕ=arcsin(3/A)=arccos(4/A)\phi=arcsin(3/A)=arccos(4/A)

so, the resultant oscillation:


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

second option:


x2(t)=3sin(ωtπ/2)=3cos(ωt)x_2​(t)=3sin(ωt-π/2)=-3cos(ωt)A=(3)2+42=5A=\sqrt{(-3)^2+4^2}=5ϕ=arcsin(3/A)=arcsin(3/A)\phi=arcsin(-3/A)=-arcsin(3/A)

then, the resultant oscillation:


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


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