Question

A wave disturbance in a medium is described by $y(x,t) = 0.02\cos (50\pi t + \dfrac{\pi }{2})\cos (10\pi x)$ where $x$and $y$are in meter and $t$is in seconds. Which of the following is correct?
A. A node occurs at $x = 0.15m$
B. An antinode occurs at $x = 0.3m$
C. The speed wave is $5m{s^{ - 1}}$
D. The wavelength is $0.3m$

Answer 1

Hint: To solve the problem first compare the given equation with the generalized wave equation and then use the necessary conditions to find the correct answer.

Formula used: The generalized wave equation: $y(x,t) = A\cos (\omega t + \dfrac{\pi }{2})\cos (kx)$;
where $A$represents amplitude, $\omega $and $k$are constants, $x$represents displacement in meters and $t$represents time in seconds.
Wave speed$(v) = \dfrac{\omega }{k}$; Wavelength$(\lambda ) = \dfrac{{2\pi }}{\lambda }$;

Complete step by step answer:
In the question we have a wave equation and we are asked to choose the correct option from the given ones.
The given wave equation is,
$y(x,t) = 0.02\cos (50\pi t + \dfrac{\pi }{2})\cos (10\pi x)$; where $x$and $y$are in meter and $t$is in seconds.
To solve this problem, we will compare the given wave equation with the generalized wave equation.
The generalized wave equation can be represented as:
$y(x,t) = A\cos (\omega t + \dfrac{\pi }{2})\cos (kx)$
So, after comparison we get: $A = 0.02$; $\omega = 50\pi $;$k = 10\pi $
Now, for the node to occur we must have $y = 0$. This is possible only when $\cos kx = 0$. So, we can write that
$\cos kx = \cos \dfrac{\pi }{2}$
$ \Rightarrow kx = \dfrac{\pi }{2}$
$ \Rightarrow x = \dfrac{\pi }{{2k}}$
Now substituting the value of $k$in the previous equation we have:
$ \Rightarrow x = \dfrac{\pi }{{2 \times 10\pi }} = \dfrac{1}{{20}}m = 0.05m$
Thus, we get that the node occurs at $x = 0.05m$. Hence the first option is incorrect.
For antinode the displacement of $y$must be maximum and hence the condition for checking is
$kx = \pi $
$ \Rightarrow x = \dfrac{\pi }{k}$
Substituting the value of $k$in the previous equation we have:
$ \Rightarrow x = \dfrac{\pi }{{10\pi }}$
$ \Rightarrow x = 0.1m$
Thus, an antinode occurs at $x = 0.1m$. So, the second option is also incorrect.
Now, let us calculate the wave speed. We know that wave speed is calculated using the formula
$v = \dfrac{\omega }{k}$
Thus, substituting the values of $\omega $and $k$ respectively in the previous equation we have:
$v = \dfrac{{50\pi }}{{10\pi }}m{s^{ - 1}} = 5m{s^{ - 1}}$
So, we got that the wave speed is equal to $5m{s^{ - 1}}$and hence option C is correct.
Now let us check what would be the wavelength. We know,
$k = \dfrac{{2\pi }}{\lambda }$
So,
$\lambda = \dfrac{{2\pi }}{k}$
Substituting the value of $k$in the previous equation we have:
$\lambda = \dfrac{{2\pi }}{{10\pi }}m = 0.2m$
Thus, we get the wavelength as $0.2m$ and hence option D is not correct.
Hence, the correct answer is option C.

Note: In general, the wave equation is written as: $y(x,t) = A\sin \omega t\cos kx$; where all the terms have the same physical meaning as stated above. As $\sin \omega t$can be rewritten as $\cos (\omega t + \dfrac{\pi }{2})$ here , due to the necessity of the problem we have written the generalized wave equation as $y(x,t) = A\cos (\omega t + \dfrac{\pi }{2})\cos (kx)$