Question

A wave has SHM whose period is $4s$ while another wave which also possesses SHM has its period $3s$. If both are combined, then the resultant wave will have the period equal to
$A)4s$
$B)5s$
$C)12s$
$D)3.43s$

Answer 1

A wave is said to have simple harmonic motion (SHM), if the particle which generates this wave undergoes simple harmonic motion. Such waves have frequencies, which are nothing but the reciprocal of their time periods. When two waves having SHM superimpose, the concept of beat arises.

Complete step by step answer:
When two waves having simple harmonic motion interfere, both constructive interference as well as destructive interference occur. If the frequencies or time periods of these waves are comparable, then the frequency of resultant wave is termed as beat frequency.
We know that beats are related to sound waves. When two waves of different frequencies or time periods combine together, a sensation of beat arises, in which we hear or sense sound, in a particular pattern and rhythm. The frequency of resultant waves is called the beat frequency, which is equal to the difference in frequencies of individual waves. For example, let us consider two waves $A$ and $B$,of frequencies ${{f}_{A}}$ and ${{f}_{B}}$, respectively. We know that the time period of a wave is equal to the reciprocal of its frequency. Therefore, time periods of $A$ and $B$ are given by
${{T}_{A}}=\dfrac{1}{{{f}_{A}}}$
and
${{T}_{B}}=\dfrac{1}{{{f}_{B}}}$
where,
${{T}_{A}}$ and ${{T}_{B}}$ are the time periods of wave $A$ and wave $B$, respectively.
Let this set of equations be represented by X.
Now, when waves $A$ and $B$ superimpose, the frequency of resultant wave is known as beat frequency and is given by
${{f}_{b}}={{f}_{A}}-{{f}_{B}}$
where
${{f}_{b}}$ is the beat frequency of resultant wave
Let this be equation Y.
If ${{T}_{b}}$ is the time period of resultant wave, it is given by
$\dfrac{1}{{{T}_{b}}}=\dfrac{1}{{{T}_{A}}}-\dfrac{1}{{{T}_{B}}}$
Let this be equation Z.
${{T}_{b}}$ or time period of resultant wave refers to the time required for one beat while ${{f}_{b}}$ or frequency of resultant wave refers to the number of beats in one second.
Coming to the question, we are provided that two waves of time periods $3s$ and $4s$, combine together, to form a resultant wave. From equation Z, we know that time period of the resultant wave would be the time period of beats formed, and is given by:
$\dfrac{1}{{{T}_{b}}}=\dfrac{1}{{{T}_{A}}}-\dfrac{1}{{{T}_{B}}}\Rightarrow \dfrac{1}{{{T}_{b}}}=\dfrac{1}{3}-\dfrac{1}{4}\Rightarrow {{T}_{b}}=12s$
Therefore, the time period of the resultant wave is equal to $12s$ and the correct answer is option $C$.

Note:
When two waves of different frequencies superimpose, the beat frequency of the resultant wave is equal to the difference in frequencies of individual waves. Here, students need to understand that beat frequency is equal to the magnitude of difference between two frequencies. Therefore, equation Y can be clearly written as
${{f}_{b}}={{f}_{A}}-{{f}_{B}}={{f}_{B}}-{{f}_{A}}=\left| {{f}_{A}}-{{f}_{B}} \right|$