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

Solution

Physics sir · 4 steps

Split the product into two cosines, then the beat period is 2π over the difference of their ω values.

1. Find the two waves. Use 2 cos A cos B = cos(A+B) + cos(A-B) with A = 50.5t and B = 1.5t. The result is the sum of the two original waves.

   x = a/2[cos(52t) + cos(49t)]

2. Beat frequency. The two angular frequencies are ω₁ = 52 rad s⁻¹ and ω₂ = 49 rad s⁻¹. The beat frequency is ν_b = (ω₁ - ω₂)/2π = 3/2π Hz.
3. Beat period. T_b = 1/ν_b = 2π/3 ≈ 2.09 s. This matches the envelope cos(1.5t): the loudness depends on its magnitude, which repeats every π/1.5 ≈ 2.09 s.
4. So. The waves beat with a period of about 2 s, which is option (c).

The idea to keep. Beat frequency is the difference of the two frequencies. The envelope term carries half that difference, so its own period is twice the beat period.

Answer from the NTA final answer key. Solution by PrepWiser.