AP Physics 2 Exam Question

A transverse wave travels along a rope with a speed of 2.5 m/s. The amplitude of the wave is 0.3 m, while its wavelength is 0.4 m.

1. Determine the frequency of the wave.
2. Calculate the period of the wave.
3. Find the crest-to-trough distance of the wave.
4. Calculate the wave number of the wave.
5. Determine the equation of the wave in the form of $y = A \sin(kx - \omega t)$.

Answer with Step-by-Step Explanation

1. To determine the frequency of the wave, we can use the formula $v = \lambda f$, where $v$ is the wave speed and $\lambda$ is the wavelength.

   Rearranging the formula to solve for frequency $f$, we have: $f = \frac{v}{\lambda} = \frac{2.5 \, \text{m/s}}{0.4 \, \text{m}} = 6.25 \, \text{Hz}$

   Therefore, the frequency of the wave is 6.25 Hz.

2. The period of a wave is the time it takes for one complete cycle. We can calculate it using the equation $T = \frac{1}{f}$, where $T$ is the period and $f$ is the frequency.

   Substituting the given frequency into the equation, we have: $T = \frac{1}{6.25 \, \text{Hz}} = 0.16 \, \text{s}$

   Thus, the period of the wave is 0.16 seconds.

3. The crest-to-trough distance of a wave is twice the amplitude. Therefore, the crest-to-trough distance in this case is: $2 \times 0.3 \, \text{m} = 0.6 \, \text{m}$

   Hence, the crest-to-trough distance of the wave is 0.6 meters.

4. The wave number ($k$) represents the number of wavelengths present in a given distance. It can be calculated using the formula $k = \frac{2\pi}{\lambda}$, where $\lambda$ is the wavelength.

   Substituting the given wavelength into the equation, we have: $k = \frac{2\pi}{0.4 \, \text{m}} = 15.71 \, \text{rad/m}$

   Therefore, the wave number of the wave is 15.71 rad/m.

5. The equation of a transverse wave can be expressed as $y = A \sin(kx - \omega t)$, where $y$ is the displacement of the wave, $A$ is the amplitude, $k$ is the wave number, $x$ is the position of the particle, and $\omega$ is the angular frequency.

   Since the wave is traveling in the positive x-direction, we can assume $x$ to be increasing. The general equation for a wave traveling in the positive x-direction is $y = A \sin(kx - \omega t)$.

   As the wave is transverse, the angular frequency ($\omega$) is related to the frequency ($f$) by the equation $\omega = 2\pi f$.

   Substituting the given values into the equation, we have: $y = 0.3 \, \text{m} \sin(15.71 \, \text{rad/m} \cdot x - 2\pi \cdot 6.25 \, \text{Hz} \cdot t)$

   Thus, the equation of the given wave is $y = 0.3 \sin(15.71x - 39.27t)$.