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Physics Question 34 – JEE-MAIN 2025

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)

When two waves of slightly different frequencies superimpose, they produce beats, which are periodic variations in the amplitude of the resultant wave.

🥷
Ninja StrategyEstimate Beat Period

Recognize that the beat angular frequency is 2×1.5=3 rad/s. The beat period is 2π/3, which is approximately 6.28/32.09 seconds. This quickly points to 2s as the closest integer.

Step 1: Relate to Standard Beat Equation✦ Active

The given wave equation is x=acos(1.5t)cos(50.5t). This form is equivalent to the superposition of two waves y=A0cos(ω1t)+A0cos(ω2t), which results in y=2A0cos(ω1+ω22t)cos(ω1ω22t). Comparing the given equation with this standard form:

ω1ω22=1.5ω1ω2=3 rad/s
ω1+ω22=50.5ω1+ω2=101 rad/s
Step 2: Calculate Beat Angular Frequency and Period○ Expand

The angular beat frequency, ωbeat, is the absolute difference between the angular frequencies of the two component waves.

ωbeat=|ω1ω2|=3 rad/s

The beat period, Tbeat, is given by Tbeat=2πωbeat.

Tbeat=2π3 s
💡 Teacher's Secret Hint

Remember that the beat period is the reciprocal of the beat frequency, and for angular frequencies, it's 2π divided by the angular beat frequency.

Step 3: Calculate Numerical Value and Round○ Expand

Substitute the value of π3.14159 into the expression for Tbeat:

Tbeat=2×3.1415936.2831832.094 s

Rounding to the nearest integer, the beat period is 2 s.

💡 Teacher's Secret Hint

Always pay attention to rounding instructions, especially 'close to nearest integer'.

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