transverse periodic wave is represented by the equation $\mathrm{z}(\mathrm{y}, \mathrm{t})=1.50 \mathrm{~cm} \sin \left(1,250 \mathrm{rad} / \mathrm{s} \mathrm{t}+10.0 \mathrm{~m}^{-1} \mathrm{y}\right)$. What is the direction of the vibration of the wave?
A) the $y$ direction
B) the $x$ direction
C) the $z$ direction

Question

Quizplus · Accepted Answer

In a transverse wave, the direction of vibration is perpendicular to the direction of wave propagation. The given equation $z(y, t)=1.50 \, \text{cm} \sin(1,250 \, \text{rad/s} \, t+10.0 \, \text{m}^{-1} \, y)$ indicates that the wave is propagating in the $y$ direction (as $y$ appears in the argument of the sine function) and the displacement of the wave is in the $z$ direction (since the function is $z(y, t)$). Therefore, the vibration of the wave is in the $z$ direction.

Transverse Periodic Wave Is Represented by the Equation z(y,t)=1.50 cmsin⁡(1,250rad/st+10.0 m−1y)\mathrm{z}(\mathrm{y}, \mathrm{t})=1.50 \mathrm{~cm} \sin \left(1,250 \mathrm{rad} / \mathrm{s} \mathrm{t}+10.0 \mathrm{~m}^{-1} \mathrm{y}\right)z(y,t)=1.50 cmsin(1,250rad/st+10.0 m−1y)

Transverse Periodic Wave Is Represented by the Equation $\mathrm{z}(\mathrm{y}, \mathrm{t})=1.50 \mathrm{~cm} \sin \left(1,250 \mathrm{rad} / \mathrm{s} \mathrm{t}+10.0 \mathrm{~m}^{-1} \mathrm{y}\right)$