Question

A transverse wave propagating along the x−axis is represented by y(x,t)=8.0sin(0.5πx−4πt−π/4),where x is in meters and t is in seconds. The speed of the wave is
A 8m/s
B 4m/s
C 0.5m/s
D 3.14/4m/s

Answer 1

Hint:- Analyze the given equation of transverse wave along x-axis and compare it with general equation to find the required wave number, amplitude, angular velocity and then with the help of expression of relationship between wavelength and time period to find the required velocity

Formula used
General equation of wave
\[y(x,t) = Asin(kx - \omega t + \phi )\]
Where A is amplitude, $\omega $is angular velocity ,$\phi $ is phase difference\[\
  :{\text{ }}, \\
    \\
\\]
\[v = \dfrac{\lambda }{T}\]
Where $\lambda $ is wavelength

Complete step-by-step solution:
We know that general equation of wave is
\[y(x,t) = Asin(kx - \omega t + \phi )\] (1)
And we have given given equation of transverse wave along x-axis is
\[y(x,t) = 8sin(0.5\pi x - 4\pi t - \pi /4\] (2)
Comparing equation (1) and (2) we get,
The quantity \[2\pi /\lambda \], which occurs in the mathematical description of wave motion, is called the wave number k
\[k = 2\pi /\lambda = 0.5\pi \]
\[\lambda = 4m\] (3)
And angular velocity Angular velocity is the rate of change of angular displacement and angular acceleration is the rate of change of angular velocity.
\[\omega = \dfrac{{2\pi }}{T} = 4\pi \]
So time period will be
\[T = 0.5s\] (4)
Now we know that velocity is equal to wavelength by time period
\[v = \dfrac{\lambda }{T}\]
Then from equation (3) and (4)
\[\
  v = \dfrac{4}{{0.5}} \\
   \Rightarrow v = 8m/s \\
\ \]
Hence option A is correct

Note:- The speed of a wave is proportional to the wavelength and indirectly proportional to the period of the wave: \[v = \lambda T{\text{ }}v{\text{ }} = {\text{ }}\lambda {\text{ }}T\]
 The type of wave that occurs in a string is called a transverse wave. This equation can be simplified by using the relationship between frequency and period, \[v = \lambda f\]