How Do Phase Differences and Velocity Relate in Wave Propagation?

[timtng]

A transverse wave of frequency 40 Hz
propagates down a string. Two points 5 cm apart are
out of phase by p/6. (a) What is the wavelength of the
wave? (b) At a given point, what is the phase
difference between two displacements for times 5 ms
apart? (c) What is the wave velocity?

for a.) I use theta=(sπx)/λ
solving for λ I get λ=.6m

Please help me on part b and c, and check to see if I did part a correctly.

 

[chroot]

For part b, you can figure out the phase difference by looking at the frequency of the wave. The frequency is 40 Hz, so the wave has a period T = 1/40 Hz = 0.025 s. The time period 50 ms is thus 0.050 / 0.025 = 2 periods, exactly. If the point on the string executes exactly an integer number of periods in 50 ms, then its phase difference between the beginning and end of that 50 ms period is zero.

For part c, you know the frequency, 40 Hz, and the wavelength, 0.6 m. You can find the velocity with

$$v = \lambda \cdot \nu$$

Does this make sense?

- Warren

 

[M98Ranger]

b) The phase difference between two displacements at times 5 ms apart can be found by first calculating the period of the wave, which is equal to 1/frequency. In this case, the period is 1/40 Hz = 0.025 seconds. Then, we can use the equation theta = (2πft) to find the phase difference, where f is the frequency and t is the time difference. Plugging in the values, we get theta = (2π*40 Hz*0.005 s) = 0.4π radians. This means that the two displacements are out of phase by 0.4π radians or 72 degrees.

c) The wave velocity can be calculated using the equation v = λf, where v is the velocity, λ is the wavelength, and f is the frequency. Plugging in the values, we get v = 0.6m * 40 Hz = 24 m/s. Therefore, the wave velocity is 24 m/s.

Your calculation for part a looks correct. Just a minor note, the wavelength is usually represented by the Greek letter lambda (λ), not the letter l.