Related Formula
The product of cosines can be transformed into a sum using the trigonometric identity:
A B = (1)/(2) [ (A + B) + (A - B)]$$\cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)]$$
The beat frequency fbeat$f_{\text{beat}}$ is given by:
fbeat = |f₁ - f₂| = | (ω₁ - ω₂)/(2π) |$$f_{\text{beat}} = |f_1 - f_2| = \left| \frac{\omega_1 - \omega_2}{2\pi} \right|$$
The beat period Tbeat$T_{\text{beat}}$ is:
Tbeat = 1fbeat$$T_{\text{beat}} = \frac{1}{f_{\text{beat}}}$$
Core Logic
Rewrite the superposition equation:
x = a (1.5t) (50.5t)$$x = a \cos(1.5t) \cos(50.5t)$$
Apply the identity with A = 50.5t$A = 50.5t$ and B = 1.5t$B = 1.5t$:
x = (a)/(2) [ (52t) + (49t)]$$x = \frac{a}{2} [\cos(52t) + \cos(49t)]$$
Here, the two component frequencies are:
ω₁ = 52 ~rad/s f₁ = (52)/(2π)$$\omega_1 = 52 \mathrm{~rad/s} \implies f_1 = \frac{52}{2\pi}$$
ω₂ = 49 ~rad/s f₂ = (49)/(2π)$$\omega_2 = 49 \mathrm{~rad/s} \implies f_2 = \frac{49}{2\pi}$$
Calculate the beat frequency:
fbeat = f₁ - f₂ = (52 - 49)/(2π) = (3)/(2π) ~Hz$$f_{\text{beat}} = f_1 - f_2 = \frac{52 - 49}{2\pi} = \frac{3}{2\pi} \mathrm{~Hz}$$
Step 1: Calculate Beat Period
The time period of beats is:
Tbeat = 1fbeat = (2π)/(3) ≈ (2 × 3.14)/(3) = 2.09 ~s$$T_{\text{beat}} = \frac{1}{f_{\text{beat}}} = \frac{2\pi}{3} \approx \frac{2 \times 3.14}{3} = 2.09 \mathrm{~s}$$
Rounding to the nearest integer gives 2 ~s$2 \mathrm{~s}$.
Pattern Recognition
Sees: product of two cosines with significantly different coefficients ω₁$\omega_1$ and ω₂$\omega_2$.
Shortcut: The beat period is simply 2π$2\pi$ divided by the difference between the two component frequencies, where the component frequencies are (ωaverage ± ωenvelope)$(\omega_{\text{average}} \pm \omega_{\text{envelope}})$. The difference is 2 × ωenvelope = 2 × 1.5 = 3 ~rad/s$2 \times \omega_{\text{envelope}} = 2 \times 1.5 = 3 \mathrm{~rad/s}$. Thus, T = 2π / 3 ≈ 2 ~s$T = 2\pi / 3 \approx 2 \mathrm{~s}$.
Chapter Mix
Class 11 Physics: Waves
