Two harmonic waves moving in the same direction superimpose to form a wave x= acos ⁡ (1.5t ) cos ⁡ (50.5t ) where t is in seconds. Find the period with which they beat. (close to nearest integer)

Question

Accepted Answer

Correct answer: C. (c) The superposition of the two harmonic waves results in the wave equation: x= acos ⁡ (1.5t ) cos ⁡ (50.5t ) This equation can be expanded using the trigonometric identity for the product of cosines: x= a2 cos ⁡ [ (1.5+50.5 )t ]+ a2 cos ⁡ [ (50.5-1.5 )t ] Simplifying, we find: x= a2 cos ⁡ (52t )+ a2 cos ⁡ (49t ) This represents two waves with angular frequencies 52 and 49, respectively. To find the beat frequency, we calculate the differences in frequencies: f_1= 522 ,f_2= 492 The beat frequency is then: f_ Beat=f_1-f_2= 32 Hz The period of the beats, T_ Beat, is the reciprocal of the beat frequency: T_ Beat= 1f_ Beat= 2 3 sec Approximating this gives: T_ Beat ≈2.09 sec ≈2 sec

$Two harmonic waves moving in the same direction superimpose to form a wave x = acos (1.5 t) cos (50.5 t) where t is in seconds. Find the period with which they beat. (close to nearest integer)$

Solution

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