Question

When a stationary wave is formed, then its frequency is
A. Same as that of the individual waves
B. Twice that of the individual waves
C. Half that of the individual waves
D. That of the individual waves

Answer 1

Using the principle of stationary waves, first write the equation of two different stationary waves one is moving in positive X- axis and the other moving in negative X-axis having dame time period wavelength and amplitude. After combining two waves using the superposition principle we can get the frequency of the resultant stationary wave which helps us to choose the correct option.

Complete step by step answer:
Stationary wave is also known as a standing wave which is a combination of two waves moving in opposite directions, each having the same amplitude and frequency. Equation for stationary wave is given as:
$y = A\sin \left( {2\pi \left[ {ft \pm \dfrac{x}{\lambda }} \right]} \right) \ldots \ldots \left( 1 \right)$
$\Rightarrow f = \dfrac{1}{T} \ldots \ldots \left( 2 \right)$

Now, with the help of above definition, let us consider a progreesive wave having amplitude $A$ ,wavelength $\lambda$ and frequency $f$ travelling in the positive X-axis direction
With the help of equation $\left( 1 \right)$ and $\left( 2 \right)$ we will get,
$y_1 = A\sin \left( {2\pi \left[ {ft - \dfrac{x}{\lambda }} \right]} \right)$
Here $f = \dfrac{1}{T}$
$y_1 = A\sin \left( {2\pi \left[ {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right]} \right) \ldots \ldots \left( 3 \right)$
This wave reflected from the free end and now it travels in negative X-axis direction with same amplitude $A$ ,wavelength $\lambda$ and frequency $f$
With the help of equation $\left( 1 \right)$ and $\left( 2 \right)$ we will get,
$y_2 = A\sin \left( {2\pi \left[ {ft + \dfrac{x}{\lambda }} \right]} \right)$
Here $f = \dfrac{1}{T}$
$y_2 = A\sin \left( {2\pi \left[ {\dfrac{t}{T} + \dfrac{x}{\lambda }} \right]} \right) \ldots \ldots \left( 4 \right)$

Using principle of superposition to get the resultant stationary wave we will get,
Adding equation $\left( 3 \right)$ and $\left( 4 \right)$ we will get,
$y = y_1 + y_2$
$\Rightarrow y = A\sin \left( {2\pi \left[ {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right]} \right) + A\sin \left( {2\pi \left[ {\dfrac{t}{T} + \dfrac{x}{\lambda }} \right]} \right)$
Taking $A$ as common terms we will get,
$\Rightarrow y = A\left( {\sin \left( {2\pi \left[ {\dfrac{t}{T} - \dfrac{x}{\lambda }} \right]} \right) + \sin \left( {2\pi \left[ {\dfrac{t}{T} + \dfrac{x}{\lambda }} \right]} \right)} \right)$
Using $\sin A + \sin B = 2\sin \left( {\dfrac{{a + b}}{2}} \right)\cos \left( {\dfrac{{a - b}}{2}} \right)$ formula we will get,
$\Rightarrow y = A\left( {2\sin 2\pi \left( {\dfrac{{\dfrac{t}{T} - \dfrac{x}{\lambda } + \dfrac{t}{T} + \dfrac{x}{\lambda }}}{2}} \right)\cos 2\pi \left( {\dfrac{{\dfrac{t}{T} - \dfrac{x}{\lambda } - \dfrac{t}{T} - \dfrac{x}{\lambda }}}{2}} \right)} \right)$

Further simplifying we will get,
$y = A\left( {2\sin 2\pi \left( {\dfrac{{\dfrac{t}{T} + \dfrac{t}{T}}}{2}} \right)\cos 2\pi \left( {\dfrac{{ - \dfrac{x}{\lambda } - \dfrac{x}{\lambda }}}{2}} \right)} \right)$
$\Rightarrow y = A\left( {2\sin 2\pi \left( {\dfrac{{\dfrac{{2t}}{T}}}{2}} \right)\cos 2\pi \left( {\dfrac{{ - \dfrac{{2x}}{\lambda }}}{2}} \right)} \right)$
$\Rightarrow y = A\left( {2\sin 2\pi \left( {\dfrac{t}{T}} \right)\cos 2\pi \left( {\dfrac{{ - x}}{\lambda }} \right)} \right)$
We know $\cos \left( { - \theta } \right) = \cos \theta$
Hence,
$y = 2A\sin \left( {\dfrac{{2\pi t}}{T}} \right)\cos \left( {\dfrac{{2\pi x}}{\lambda }} \right)$
Here we can see that the frequency of the resultat stationary wave is also the same as that of the other two waves, that is $\dfrac{1}{T} = f$.

Therefore the correct option is A.

Note: Always keep in mind before finding the resultant stationary wave make sure that the individual stationary waves are of the same time period, wavelength and amplitude.The individual waves must come from the same wave but in opposite directions.