Question

The equation of a wave disturbance is given as $y=0.02\cos \left( 50\pi t+\dfrac{a}{2} \right)\cos (10\pi x)$ where x and y are in meters and t is in second. Choose the correct option
a) Antinodes occurs at x=0.3m
b) The wavelength is 0.2m
c) The speed of the constituent wave is 4m/s
d) Nodes occurs at x=0.3m

Answer 1

Hint: To answer the above question we have to compare the above equation of disturbance of the wave to the general equation. By comparing the equation we would be able to determine the various parameters of the wave. Hence once their numerical values are obtained comparing the results with the options will enable us to identify the correct answer.
Formula used:
$y(x,t)=Acos(\omega t+\dfrac{\pi }{2})\cos kx$

Complete answer:
The general equation of a wave disturbance is given by,
$y(x,t)=Acos(\omega t+\dfrac{\pi }{2})\cos kx$
Where ‘A’ is the amplitude of the wave, $\omega $ is the angular frequency of the wave, ‘k’ is the propagation constant and $\dfrac{\pi }{2}$ is the phase constant.
From the above equation we can conclude that y is a function of time and x i.e. distance from a reference point. By comparing the above equation we can say that $k=10\pi $ and $\omega =50\pi $ . But we know that,
$k=\dfrac{2\pi }{\lambda }$ where $\lambda $ is the wavelength of the disturbance. Since we know that $k=10\pi $ , we get
$\begin{align}
  & k=\dfrac{2\pi }{\lambda } \\
 & \Rightarrow 10\pi =\dfrac{2\pi }{\lambda } \\
 & \Rightarrow \lambda =0.2m \\
\end{align}$

Therefore the correct answer of the above question is option b.

Note:
It is to be noted that $\omega =50\pi $ and since $\omega =2\pi \gamma $ where $\gamma $ is the frequency of the wave we get the frequency as,
$\begin{align}
  & \omega =2\pi \gamma \\
 & \Rightarrow 50\pi =2\pi \gamma \\
 & \Rightarrow \gamma =25Hz \\
\end{align}$
The speed of the wave is given by,$v=\lambda \gamma $ . Therefore the speed of the wave is,
$\begin{align}
  & v=\lambda \gamma \\
 & v=0.2m\times 25/s=5m/s \\
\end{align}$
The condition for an antinode to occur in a wave is given by $kx=0,\pi ,3\pi ....n\pi $ where n is an integer.
Similarly the condition for a node is given by $kx=\dfrac{\pi }{2},\dfrac{3\pi }{2}....\dfrac{n\pi }{2}$ where n is an integer. Clearly if we put x=0.3m, then the condition of getting an antinode is satisfied. But It is to be noted that getting a node does not solely depend on x but it depends on time as well.