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 (close to the nearest integer):

• A. 6 s
• B. 4 s
• C. 1 s
• D. 2 s

Hint:

The beat period is given by the difference between the angular frequencies of the two waves. Take care to use the correct formula and values to calculate it.

Answer 1

The Correct Option is D

Step 1: Understanding the Given Wave
The resultant wave is given by: \[ x = a \cos(1.5t) \cos(50.5t) \] This represents the product of two cosine functions, which can be interpreted as a modulated wave.
Step 2: Using Trigonometric Identity
We use the identity for the product of cosines:
\[ \cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)] \] Applying this to the given wave:
\[ x = \frac{a}{2} [\cos(1.5t + 50.5t) + \cos(50.5t - 1.5t)] \] \[ x = \frac{a}{2} [\cos(52t) + \cos(49t)] \] This shows the superposition of two waves with angular frequencies \( \omega_1 = 52 \) rad/s and \( \omega_2 = 49 \) rad/s.
Step 3: Calculating Beat Frequency
The beat frequency \( f_{\text{beat}} \) is the difference between the two frequencies: \[ f_{\text{beat}} = |f_1 - f_2| \] First, convert angular frequencies to linear frequencies: \[ f_1 = \frac{52}{2\pi} \text{ Hz}, \quad f_2 = \frac{49}{2\pi} \text{ Hz} \] Thus: \[ f_{\text{beat}} = \frac{|52 - 49|}{2\pi} = \frac{3}{2\pi} \text{ Hz} \]
Step 4: Calculating Beat Period
The beat period \( T_{\text{beat}} \) is the inverse of the beat frequency: \[ T_{\text{beat}} = \frac{1}{f_{\text{beat}}} = \frac{2\pi}{3} \text{ seconds} \] Numerically: \[ T_{\text{beat}} \approx \frac{6.2832}{3} \approx 2.0944 \text{ s} \]
Step 5: Matching with Options
The closest integer to 2.0944 s is 2 s, which corresponds to option (4).
Final Answer
The correct answer is 4.