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 $ .