Question #159650

A sinusoidal wave travelling in the positive x direction has an amllplitude of 15.0 cm, a wavelength of 40.0 cm, and a frequency pf 8.00 Hz. The vertical position of an element of the medium at t=0 and x =0 is also 15.0 cm.

A. Find the wave number k, period T, angular frequency w, and speed v of the wave

B. Write a general expression for the wave function


1
Expert's answer
2021-01-29T20:01:39-0500

Let's denote: A=15cm=0.15mA = 15cm =0.15m is the amplitude, λ=40cm=0.4m\lambda = 40cm = 0.4m is the wavelength, f=8Hzf = 8Hz is the frequency.

The wave number is given as:


k=2πλ=2π0.4=5π m1k = \dfrac{2\pi}{\lambda} = \dfrac{2\pi}{0.4} = 5\pi\space m^{-1}

The angular frequency is:


ω=2πf=16π rad/s\omega = 2\pi f = 16\pi \space rad/s

The speed is:


v=λf=3.2m/sv = \lambda f = 3.2m/s

The general expression for the wave is:


y(x,t)=Acos(kxωt)y(x,t)=0.15cos(5πx16πt) [m]y(x,t) = A\cos(kx - \omega t)\\ y(x,t) = 0.15\cos(5\pi x - 16\pi t)\space [m]

Answer. A) k=5π m1,ω=16π rad/s,v=3.2m/sk = 5\pi \space m^{-1}, \omega = 16\pi \space rad/s, v = 3.2m/s, B) y(x,t)=0.15cos(5πx16πt) [m]y(x,t) = 0.15\cos(5\pi x - 16\pi t)\space [m].


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