transverse periodic wave is represented by the equation $y(x, t)=2.50 \mathrm{~cm} \cos \left(2,500 \mathrm{rad} / \mathrm{st}-15.0 \mathrm{~m}^{-1} \mathrm{x}\right)$. What is the frequency of the vibration of the wave?
A) $398 \mathrm{~Hz}$
B) $422 \mathrm{~Hz}$
C) $302 \mathrm{~Hz}$
D) $490 \mathrm{~Hz}$
E) $467 \mathrm{~Hz}$

Question

Quizplus · Accepted Answer

The angular frequency $\omega$ is given as $2,500 \mathrm{~rad/s}$. The frequency $f$ is $\omega/2\pi$, which calculates to approximately $398 \mathrm{~Hz}$.

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