Question

The superposition of two SHMs of the same direction results in the oscillation of a point according to the law \[x = {x_0}\cos (2.1t)\cos (50t)\] Find the angular frequencies of the constituent oscillations and period with which they beat.
A) 52.1s ,47.98s ,0.2s
B) 50s .2.1s .0.22s
C) 52.1s 1,47.9s ,1.5s
D) none

Answer 1

Hint: In this question we are given the resultant wave produced as a result of superposition of 2 independent waves. First we need to find those 2 waves and then we will find the angular frequencies of the 2 waves. Further, we will find the frequency of the beat produced. From that we will evaluate the time period of the beat.

Complete step by step solution
The resultant of the 2 wave is given as: \[x = {x_0}\cos (2.1t)\cos (50t)\]
Using trigonometric identity to separate the cosine terms, we get: \[x = {x_0}\cos (2.1t)\cos (50t) = \dfrac{1}{2}{x_0}[\cos 52.1t + \cos 47.9t]\]
From the above equation we can clearly see that the 2 waves are \[\dfrac{1}{2}{x_0}\cos 52.1t\] and \[\dfrac{1}{2}{x_0}\cos 47.9t\] . The angular frequencies of these waves are \[52.1{s^{ - 1}}\] and \[47.9{s^{ - 1}}\] .
Now, the formula for frequency of beats is given as \[f = \dfrac{{{\omega _1} - {\omega _2}}}{{2\pi }}\]
 \[f = \dfrac{{52.1 - 47.9}}{{2\pi }} = \dfrac{{4.2}}{{2\pi }}\]
Now we know that the time period is the reciprocal of the frequency, so:
 \[T = \dfrac{1}{f} = \dfrac{{2\pi }}{{4.2}} = 1.5s\]

Therefore, the correct answer is option C

Note
Whenever you get a resultant wave, you need to break it into 2 or more different waves having cosine terms, sine terms or both. This can be done by manipulating the equation using trigonometric identities.