transverse periodic wave is represented by the equation $\mathrm{z}(\mathrm{y}, \mathrm{t})=1.50 \mathrm{~cm} \sin \left(1,250 \mathrm{rad} / \mathrm{s} t+10.0 \mathrm{~m}^{-1} \mathrm{y}\right)$. What is the frequency of the vibration of the wave?
A) $240 \mathrm{~Hz}$
B) $199 \mathrm{~Hz}$
C) $289 \mathrm{~Hz}$
D) $319 \mathrm{~Hz}$
E) $150 \mathrm{~Hz}$

Question

Quizplus · Accepted Answer

The frequency $f$ of the wave can be determined from the angular frequency $\omega$, which is given in the equation as $1,250 \, \text{rad/s}$. The relationship between angular frequency and frequency is $\omega = 2\pi f$. Solving for $f$, we get $f = \frac{\omega}{2\pi} = \frac{1250}{2\pi} \approx 199 \, \text{Hz}$.

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} 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 $\mathrm{z}(\mathrm{y}, \mathrm{t})=1.50 \mathrm{~cm} \sin \left(1,250 \mathrm{rad} / \mathrm{s} t+10.0 \mathrm{~m}^{-1} \mathrm{y}\right)$