Tag: beats in sound waves

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Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

If 2 waves of same frequency and same amplitude on superposition, produce a resultant disturbance of the same amplitude, the waves differ in phase by

  1. $ \dfrac { \pi }{ 3 } $
  2. $ \dfrac { 2\pi }{ 3 } $
  3. $ { \pi } $
  4. $ { 3\pi } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the amplitude of each wave be $A$ i.e.  $A _1 = A _2 = A$
Thus amplitude of resultant wave  $A _R =A$
Using   $A^2 _R =  A _1^2 + A _2^2 + 2A _1A _2 \cos\theta$
$\therefore$  $A^2 =  A^2 + A^2 + 2A^2 \cos\theta$
Or  $\cos\theta = -0.5$
$\implies  \ \theta = \dfrac{2\pi}{3}$

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

Two identical wires are stretched by the same tension of $101$ N & each emits a note of frequency $202$ Hz. If the tension in one wire is increased by $1N$ , then the beat frequency is:

  1. $2 Hz$
  2. $\frac{1}{2}Hz$
  3. $1 Hz$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Frequency f is proportional to sqrt(T). f1 = 202 Hz at T1 = 101 N. New tension T2 = 102 N. f2 = f1 * sqrt(T2 / T1) = 202 * sqrt(102 / 101) = 202 * sqrt(1 + 1/101) approx 202 * (1 + 1/202) = 202 + 1 = 203 Hz. Beat frequency = 203 - 202 = 1 Hz.