Question

The equation of a transverse wave propagating on a string is given by

TT y(x,t)=0.08 sen 107 x-67t + 2

a) Show that this equation can be written as:

b) Find your amplitude, wave number, angular frequency and indicate in which direction the disturbance is propagated.

c) Find the propagation speed, wavelength, frequency, and wave period.

d) Find and draw the wave profile between x=0 and x=90cm, in t=0s.

e)Write the equation that defines the motion of the point x = 10cm at every instant of
weather. Find the speed and vertical acceleration of this point at all times.

f) Show that the motion described in letter e) satisfies the differential equation of a
simple harmonic motion.

Answer 1

The given wave is

y = 0.08 sin(1072 67t + 1/2)

It is of the form

$y=A \sin(\theta+\pi/2)$

From basic trigonometry,

$\sin(\theta+\pi/2)=\cos(\theta)$

Thus, the given wave is:

$y=0.08 \cos(10\pi x -6\pi t)$

The amplitude is NOT 0.04. It is 0.08.

The standard form of a wave traveling in the positive x direction is:

$y=A \sin(kx-\omega t+\phi)$

Where k is the wave number, omega is the angular velocity and phi is the phase.

On comparing with the given wave equation, we have:

A= 0.08 m,

k=$10\pi$ m^-1

$\omega=6\pi$ rad/s

The wavelength is given by:

$k=\frac{2\pi}{\lambda}\implies \lambda=\frac{2\pi}{k}=\frac{2\pi}{10\pi}=0.2\text{ m}$

The frequency is given by:

$\omega=2\pi \nu\implies \nu = \frac{\omega}{2\pi}=\frac{6\pi}{2\pi}=3\text{ s}^{-1}$

The time period of the wave is:

$T=\frac{1}{\nu}=\frac{1}{3}=0.33 \text{ s}$

The wave is

y = 0.08 sin(1072 67t + 1/2)

at x=10 cm =0.1 m, we have:

$y=0.08 \sin(10\pi 0.1 -6\pi t+\pi/2)=0.08 \sin(\pi -6\pi t+\pi/2)$

differentiating this expression with respect to time, we get te velocity, v. Thus,

$v=\frac{dy}{dt}=0.08 \cos(\pi -6\pi t+\pi/2)\times(-6\pi)=-1.5\cos(\pi -6\pi t+\pi/2)$

again differentiating the velocity, we get the acceleration, a. Thus,

$a=\frac{dv}{dt}=-1.5\times -\sin(\pi -6\pi t+\pi/2)\times(-6\pi) \newline\newline =-28.42\sin(\pi -6\pi t+\pi/2)$

Since the acceleration of any point is

$a=-28.42\sin(10\pi x -6\pi t+\pi/2)$

which is the position of the point times a constant, it is a solution to the wave equation which says that

$\frac{d^2 x}{dt^2}=-k^2 x$

or

$a=-kx$

which is what our analysis tells us too.