Question

The stationary wave produced on a string is represented by the equation $y = 5cos\left( {\pi x/3} \right)sin\left( {40\pi t} \right)\;$ where $x$ and $y$ are in cm and $t$ is in s. The distance between consecutive nodes is:
A) 5 cm
B) $\pi \,cm$
C) 3 cm
D) 40 cm

Answer 1

Hint : Standing waves are generated when two waves on a string superimpose to form standing wave patterns on a string. In a standing wave, there are nodes which are regions of zero amplitude, and antinodes which are regions of maximum amplitude.

Formula used: The equation of a standing wave is given as,
$y = 2A\sin (kx)\cos (\omega t)$ where $y$ is the displacement of a point on the string, $k$ is the wave constant, $x$ is the displacement of the particle from the reference point, $\omega$ is the angular frequency of the waves, and $t$ is the time

Complete step by step answer
We’ve been given the equation of waves produced on a string as
$y = 5cos\left( {\pi x/3} \right)sin\left( {40\pi t} \right)\;$
Let’s compare it with the standard equation of standing waves on a string,
$y = 2A\sin (kx)\cos (\omega t)$
Comparing these two equations, we can write
$k = \pi /3$
Since the wave number is related to the wavelength as
$\lambda = \dfrac{{2\pi }}{k}$
We can calculate the wavelength as
$\lambda = \dfrac{{2\pi }}{{\pi /3}}$
$\Rightarrow \lambda = 6$
Now, the distance between two nodes $(x)$ for a standing wave will be equal to half of its wavelength i.e.,
$x = \lambda /2$
$\therefore x = 3\,cm$ which corresponds to option (C).

Note
We must be aware of the general form of a travelling wave on a string and also the distance between two nodes for a travelling wave which can be derived as follows:
For two nodes on a string, since the amplitude is zero for two nodes, the sines must be zero i.e.
$\sin (kx) = 0$
$\sin \left( {\dfrac{{2\pi }}{\lambda }x} \right) = 0$
Since sines are zero for integral multiples of $\pi$ , we can say
$\left( {\dfrac{{2\pi }}{\lambda }x} \right) = n\pi$
$x = \dfrac{{n\lambda }}{2}$
For two consecutive nodes,
$x = \dfrac{{(n + 1)\lambda }}{2} - \dfrac{{n\lambda }}{2}$
$\Rightarrow x = \lambda /2$
We must also not confuse the distance between two nodes $\lambda /2$ with the distance between a node and an antinode which is $\lambda /4$ .