Answer to Question #206284 in Mechanics | Relativity for KABIR

Question

• Two equations of simple harmonic motion are given as follows : x₁= 4×10⁻² cos2π(t+1/8); x₂=3×10⁻² cos2π(t+1/4) . Find the equation of the resultant oscillation?

Answer 1

Equation of SHMs are

$x₁= 4×10^{-2} cos2π(t+1/8); x₂=3×10^{-2} cos2π(t+1/4)$

Equation of resultant oscillation,

$y = x₁+x_2= 4×10^{-2} cos2π(t+1/8)+3×10^{-2} cos2π(t+1/4)$

$= [4\times 10^{-2}cos(2\pi t + \frac{\pi}{4})]+[3 \times 10^{-2}cos(2\pi t + \frac{\pi}{2})]$

Let $A = 4\times10^{-2} ; B = 3\times10^{-2}$ to make equation simple.

$y = [Acos(2\pi t + \frac{\pi}{4})]+[Bcos(2\pi t + \frac{\pi}{2})]$

Expanding both cos functions,

$y = A[cos(2\pi t )cos( \frac{\pi}{4}) - sin(2\pi t )sin( \frac{\pi}{4})]+B[cos(2\pi t)cos(\frac{\pi}{2})-sin(2\pi t)sin(\frac{\pi}{2})]$

since, $cos(\pi / 4) = sin(\pi / 4) = \frac{1}{\sqrt{2}}; sin(\pi /2) = 1;cos(\pi /2) = 0$

$y = \frac{A}{\sqrt{2}} [cos(2\pi t) - sin(2\pi t)] - Bsin(2\pi t)$

$y = \frac{A}{\sqrt{2}}cos(2\pi t) - ( \frac{A}{\sqrt{2}} + B) sin(2\pi t)$

Let $\frac{A}{\sqrt{2}} = Psin\phi; (\frac{A}{\sqrt{2}} + B) = Pcos\phi$

Then $y = Psin(\phi) cos(2\pi t) -Pcos(\phi )sin(2\pi t) = Psin(\phi - 2\pi t)$

$y = -Psin(2\pi t - \phi)$

where, $\phi = tan^{-1}\frac{\frac{A}{\sqrt{2}}}{\frac{A}{\sqrt{2}} + B} = tan^{-1}\frac{A}{A + \sqrt{2}B} = tan^{-1}\frac{4\times 10^{-2}}{4\times 10^{-2} + 3\times 10^{-2}\sqrt{2}} = 25.89^\circ$

and

$P = \sqrt{(\frac{A}{\sqrt{2}})^2 + (\frac{A}{\sqrt{2}} + B)^2} = 6.47 \times 10^{-2}$

Hence equation will be,

$y = -6.47 \times 10^{-2}sin(2\pi t - \phi)$