transverse periodic wave is represented by the equation $\mathrm{y}(\mathrm{x}, \mathrm{t})=2.50 \mathrm{~cm} \cos \left(2,500 \mathrm{rad} / \mathrm{st}-15.0 \mathrm{~m}^{-1} \mathrm{x}\right)$. What is the direction of the vibration of the wave?
A) the $\mathrm{x}$ direction
B) the $\mathrm{z}$ direction
C) the $y$ direction

Question

Quizplus · Accepted Answer

In a transverse wave, the direction of vibration is perpendicular to the direction of wave propagation. The given equation $y(x, t)=2.50 \, \text{cm} \cos(2,500 \, \text{rad/s}t-15.0 \, \text{m}^{-1}x)$ indicates that the displacement $y$ varies with both position $x$ and time $t$, meaning the vibration occurs in the $y$ direction, while the wave itself propagates along the $x$ direction.

Transverse Periodic Wave Is Represented by the Equation y(x,t)=2.50 cmcos⁡(2,500rad/st−15.0 m−1x)\mathrm{y}(\mathrm{x}, \mathrm{t})=2.50 \mathrm{~cm} \cos \left(2,500 \mathrm{rad} / \mathrm{st}-15.0 \mathrm{~m}^{-1} \mathrm{x}\right)y(x,t)=2.50 cmcos(2,500rad/st−15.0 m−1x)

Transverse Periodic Wave Is Represented by the Equation $\mathrm{y}(\mathrm{x}, \mathrm{t})=2.50 \mathrm{~cm} \cos \left(2,500 \mathrm{rad} / \mathrm{st}-15.0 \mathrm{~m}^{-1} \mathrm{x}\right)$