Question

A wave travelling along a string is described by, y(x,t)=0.005sin(80.0x-3.0t), in which the numerical constants are in SI units \[(0.005{\rm{ m, 80}}{\rm{.0 rad }}{{\rm{m}}^{ - 1}}{\rm{ and 3}}{\rm{.0 rad }}{{\rm{s}}^{ - 1}})\]. Calculate: (a) the amplitude (b) the wavelength (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time t = 20 s?

Answer 1

Hint: To solve this question we can compare the given travelling wave equation with the general form of the equation. After comparing we get the values for the related terms of the equation.






Formula used:
The general form of the sinusoidal wave is given as,
\[y(x,t) = A\sin (kx - \omega t)\]
Where A is the amplitude
k is the wavenumber
\[\omega \] is the angular frequency
x is the displacement
t is the time taken
The formula for angular frequency is given as,
\[\omega = 2\pi f\]
Where f is an ordinary frequency



Complete answer:
Travelling wave equation is given as
y(x,t)=0.005sin(80.0x-3.0t)
As the general equation of wave is
\[y(x,t) = A\sin (kx - \omega t)\]
Now comparing both the equations, we get
k=80.0, \[\omega = 3\]
(a) Amplitude, A=0.005 m = 5 mm
(b) As we know that wavelength, \[\lambda = \dfrac{{2\pi }}{k}\]
So, \[\lambda = \dfrac{{2\pi }}{{80.0}} = \dfrac{\pi }{{40}}m\]
=7.85 cm
(c)As we know \[\omega = 2\pi f\]
So, \[f = \dfrac{3}{{2\pi }} = 0.48Hz\]
Also, we know that, \[T = \dfrac{1}{f}\]
So, \[T = \dfrac{{2\pi }}{3} = 2.09\sec \]
At a distance x = 30.0 cm (or 0.3 m) and time t = 20 s,
Putting the given values in the general equation, we have
\[y(x,t) = 0.005\sin (80 \times 0.3 - 3 \times 20)\]
\[ = 0.005\sin ( - 36rad)\]
\[ = 4.95mm{\rm{ }} \approx 5mm\]








Note:A travelling wave is defined as the wave that is moving in a space. A wave which is travelling in the positive direction of the x axis can be represented by the wave equation \[y(x,t) = A\sin (kx - \omega t)\]. Here A is the amplitude and k is the propagation constant.