JEE Main Physics: Beats from cos A cos B Form 💡

 

❓ Question

Two harmonic waves moving in the same direction superimpose to form a wave:

$$x = a \cos(1.5t)\cos(50.5t)$$

(where $t$ is in seconds)

Find the period with which they beat.
(Answer close to nearest integer)

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🖼️ Question Image

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✍️ Short Solution

This is a standard beats question hidden inside a product form.
Key idea:
👉 Product of cosines = amplitude modulation 🔥

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🎯 HOOK (Before Reading)

“Wave ke andar wave dikhe?
Samajh jao beats chal rahe hain!” 🎵

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🔹 Step 1 — Identify the Structure (MOST IMPORTANT 💯)**

Given:

$$x = a \cos(1.5t)\cos(50.5t)$$

This is of the form:

$$x = (\text{slow term}) \times (\text{fast term})$$

Here:

• $\cos(50.5t)$ → rapid oscillation

• $\cos(1.5t)$ → slowly varying envelope

👉 Beats occur because amplitude varies with time.

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🔹 Step 2 — Compare with Standard Identity

We know:

$$\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$$

So effectively:

$$x = \frac{a}{2} \left[ \cos(52t) + \cos(49t) \right]$$

This means two waves are present with angular frequencies:

$$\omega_1 = 52$$ $$\omega_2 = 49$$

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🔹 Step 3 — Find Beat Frequency

Beat angular frequency:

$$\Delta\omega = |\omega_1 - \omega_2| = 3$$

But envelope term was:

$$\cos(1.5t)$$

Important concept:

Envelope angular frequency = $\Delta\omega/2$

Since:

$$1.5 = \frac{3}{2}$$

So:

$$\Delta\omega = 3$$

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🔹 Step 4 — Beat Period

Beat frequency:

$$f_{\text{beat}} = \frac{\Delta\omega}{2\pi} = \frac{3}{2\pi}$$

Beat period:

$$T_{\text{beat}} = \frac{1}{f_{\text{beat}}} = \frac{2\pi}{3}$$

Now calculate numerically:

$$T_{\text{beat}} = \frac{2\pi}{3} \approx \frac{6.28}{3} \approx 2.09\text{ s}$$

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✅ Final Answer

$$\boxed{T_{\text{beat}} \approx 2\text{ seconds}}$$

(nearest integer)

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⭐ Golden JEE Insight

If wave is in form:

$$x = a\cos(\alpha t)\cos(\beta t)$$

Then:

• Fast oscillation → $\beta$

• Envelope → $\alpha$

• Beat period:

$$T = \frac{2\pi}{\Delta\omega}$$

🧠 Shortcut:

Envelope frequency determines beats