Waves Test - 6

Here, velocity = 25 cm/s & wavelength  = 100 cm

As velocity = frequency X wavelength

25 = frequency X 100 => frequency = 0.25

Hence time period = 1/frequency = 4 s

Here, frequency = 12 MHz = 12000000 Hz

Speed of wave = 300000000 m/s

Wavelength = speed/frequency = 300000000/12000000 = 25 m

Beat frequency formula for calculating the beat between two overlapping sound waves is   f_b= f₁-f₂ (f₁& f₂ are two incident sound waves).

_f1= 402 - 401 = 1 Hz

_f2 = 402 - 400 = 2 Hz

F_b = 2 - 1 = 1 Hz

Let’s start with what we are given, assign some names.
λ₀=1.00m 
λ₁=1.01m=λ₀+0.01m
We will use f₀ and f₁ for the frequencies corresponding to the two wavelength. Since the two frequencies superimposed produce a variation of 4 beats in 1 second we also have:
f₀−f₁=4
f₁=f₀−4
in general, we have v=λf and in particular:
λ₀f₀=λ₁f_1. Substituting gives:
λ₀f₀=(λ₀+0.01)(f₀−4)Simplifying gives:
0=−4λ₀+0.01f₀−0.04
we have:
0=.01f₀−4.40,so
f₀=4.40/0.01=404 . Since λ₀=1.00m we have:v=λ₀f₀
v=404 m/s

 An unknown tuning fork i is producing 4 beats/sec with the fork of frequency 256Hz. This means that their frequencies differ by 4. The unknown fork might have frequency either 252 Hz or 260 Hz.
On the application of wax, the number of beats remains to 4 per second which means they differ only by 4 and it is only possible when the unknown fork has a greater frequency.
Hence the unknown tuning fork have frequency of 260Hz.

The principle of superposition may be applied to waves  ​​​whenever two (or more) waves travelling through the same medium at the same time. The waves pass through each other without being disturbed. The net displacement of the medium at any point in space or time, is simply the sum of the individual wave displacements.

As Displacement of resultant wave is given by
y = y₁+y₂ (for same direction)
y = y₁-y₂ (for opposite direction)