A harmonic wave moving in the positive x direction has an amplitude of 3.1 cm, a speed of 37.0 cm/s, and a wavelength of 26.0 cm. Assume that the displacement is zero at x = 0 and t = 0.
Calculate the displacement (in cm) due to the wave at x = 0.0 cm, t = 2.0 s.
The wave equation for deisplacement (Y)is
Y = (3.1 cm)* sin [2 pi)(x/L - f*t)]
f is the frequency, which is
(37 cm/s)/(26 cm) = 1.423 Hz
Insert x = 0 and t = 2s to compute the answer.
I'm still not getting the correct answer when I use the above formula. When I plugged in 0 for x and 2 for t, I simplified the equation to
Y= 3.1 * sin (2pi (-2.846))
Y=3.1 * sin (-17.88)
Y=3.1 * -.307
Y=-0.952 cm
What am I doing wrong?
nevermind, I was stupid and didn't have my calculator in radians! Thank!
To calculate the displacement at a specific point (x, t) on a harmonic wave, you can use the equation:
y(x, t) = A * sin(kx - ωt)
Where:
- A is the amplitude of the wave,
- k is the wave number (2π/λ, where λ is the wavelength),
- x is the position along the x-axis,
- ω is the angular frequency (2πf, where f is the frequency),
- and t is the time.
In this case, given:
- Amplitude (A) = 3.1 cm,
- Wavelength (λ) = 26.0 cm,
- Speed (v) = 37.0 cm/s,
- Position along the x-axis (x) = 0.0 cm,
- Time (t) = 2.0 s.
First, we need to find the angular frequency (ω) using the formula ω = 2πf = 2πv/λ:
ω = 2π * 37.0 cm/s / 26.0 cm = 2π * 1.4231 ≈ 8.948 rad/s
Next, calculate the wave number (k) using the formula k = 2π/λ:
k = 2π / 26.0 cm = π / 13.0 cm ≈ 0.241 rad/cm
Now, substitute these values into the equation to find the displacement (y) at x = 0.0 cm and t = 2.0 s:
y(0.0, 2.0) = 3.1 cm * sin(0.241 rad/cm * 0.0 cm - 8.948 rad/s * 2.0 s)
= 3.1 cm * sin(-17.896 rad)
≈ 3.1 cm * sin(-2.88)
Using a calculator or trigonometric table, sin(-2.88) is approximately -0.241.
Therefore, the displacement at x = 0.0 cm and t = 2.0 s is:
y(0.0, 2.0) ≈ 3.1 cm * -0.241 = -0.7471 cm
So, the displacement at x = 0.0 cm, t = 2.0 s is approximately -0.7471 cm.