Question

A long string is tied between two posts very far apart (treat as infinitely far). Waves on this string travel with speed 71 m/s. The string is being driven by two oscillators. One is located at X 0 m and the other is located at X2 6 m. Oscillator 1 drives at 687 Hz with amplitude 1.4 cm, and oscillator 2 drives a 293 Hz with amplitude 2.9 cm. Both oscillators start at maximum displacement when t 0 s. What is the displacement in cm of the string at 1.4 m and time 4.7 s? Give you answer to the nearest 0.01 cm.

Answer 1

The wavelength of a wave is given by -

For oscillator 1, we have

$\lambda$₁ = v / f₁ = [(71 m/s) / (687 Hz)]

$\lambda$₁ = 0.103 m

For oscillator 2, we have

$\lambda$₂ = v / f₂ = [(71 m/s) / (293 Hz)]

$\lambda$₂ = 0.242 m

A wavenumber of a wave is given by -

For oscillator 1, we have

k₁ = 2$\pi$ / $\lambda$₁ = [(6.28 rad) / (0.103 m)]

k₁ = 60.9 m^-1

For oscillator 2, we have

k₂ = 2$\pi$ / $\lambda$₂ = [(6.28 rad) / (0.242 m)]

k₂ = 25.9 m^-1

An angular frequency of a wave is given by -

For oscillator 1, we have

$\omega$₁ = 2$\pi$f₁ = [(6.28 rad) (687 Hz)]

$\omega$₁ = 4314.36 rad/s

For oscillator 2, we have

$\omega$₂ = 2$\pi$f₂ = [(6.28 rad) (293 Hz)]

$\omega$₂ = 1840.04 rad/s

At t = 0 sec, both oscillators start at maximum displacement.

A wave function describing the wave which will be given as -

y (x, t) = A cos (k x + $\omega$ t)

The displacement of the string at (x=1.4 m) and time (t=4.7 s) which will be given as -

For oscillator 1, we have

y₁ = (1.4 cm) cos [(60.9 m^-1) (1.4 m) + (4314.36 rad/s) (4.7 s)]

y₁ = (1.4 cm) cos [(85.26 rad) + (20277.492 rad)]

y₁ = (1.4 cm) cos (20362.752 rad)

y₁ = [(1.4 cm) (0.4961)]

y₁ = 0.694 cm

For oscillator 2, we have

y₂ = (2.9 cm) cos [(25.9 m^-1) (1.4 m) + (1840.04 rad/s) (4.7 s)]

y₂ = (2.9 cm) cos [(36.26 rad) + (8648.188 rad)]

y₂ = (2.9 cm) cos (8684.448 rad)

y₂ = [(2.9 cm) (0.4661)]

y₂ = 1.35 cm