transverse periodic wave is represented by the equation $y(x, t)=-1.50 \mathrm{~cm} \sin \left(1,500 \mathrm{rad} / \mathrm{s} t-10.0 \mathrm{~m}^{-1}\right.$ $\mathrm{x})$. Another transverse wave is represented by the equation $\mathrm{y}(\mathrm{x}, \mathrm{t})=+1.50 \mathrm{~cm} \sin \left(1,500 \mathrm{rad} / \mathrm{s} \mathrm{t}+10.0 \mathrm{~m}^{-1}\right.$ $\mathrm{x})$. What is the equation that represents the superposition of the two waves?
A) $y(x, t)=3.0 \mathrm{~cm} \cos (1,500 \mathrm{rad} / \mathrm{st}) \sin \left(10.0 \mathrm{~m}^{-1} \mathrm{x}\right)$
B) $y(x, t)=3.0 \mathrm{~cm} \sin \left(1,500 \mathrm{rad} / \mathrm{s} t-10.0 \mathrm{~m}^{-1} \mathrm{x}\right)$
C) $y(x, t)=3.0 \mathrm{~cm} \sin (1,500 \mathrm{rad} / \mathrm{s} \mathrm{t}) \cos \left(10.0 \mathrm{~m}^{-1} \mathrm{x}\right)$
D) $y(x, t)=3.0 \mathrm{~cm} \sin \left(1,500 \mathrm{rad} / \mathrm{s} t+10.0 \mathrm{~m}^{-1} \mathrm{x}\right)$

Question

Quizplus · Accepted Answer

The superposition of the two waves can be found by using the trigonometric identity for the sum of two sine waves with opposite phase shifts in x. This results in a product of a sine and cosine function, where the amplitude is doubled and the phase shift in time is represented by a cosine function, while the spatial variation is represented by a sine function.

Transverse Periodic Wave Is Represented by the Equation y(x,t)=−1.50 cmsin⁡(1,500rad/st−10.0 m−1y(x, t)=-1.50 \mathrm{~cm} \sin \left(1,500 \mathrm{rad} / \mathrm{s} t-10.0 \mathrm{~m}^{-1}\right.y(x,t)=−1.50 cmsin(1,500rad/st−10.0 m−1

Transverse Periodic Wave Is Represented by the Equation $y(x, t)=-1.50 \mathrm{~cm} \sin \left(1,500 \mathrm{rad} / \mathrm{s} t-10.0 \mathrm{~m}^{-1}\right.$