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):
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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.
Step 1: Wave Analysis The given wave equation is:\[ x = a \cos(1.5t) \cos(50.5t) \]This equation represents the product of two cosine functions, indicating a modulated wave form. Step 2: Trigonometric Identity Application Utilizing the product-to-sum trigonometric identity for cosines:\[ \cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)] \]Applying this identity to the given wave equation yields:\[ x = \frac{a}{2} [\cos(1.5t + 50.5t) + \cos(50.5t - 1.5t)] \]\[ x = \frac{a}{2} [\cos(52t) + \cos(49t)] \]This transformed equation reveals the superposition of two individual waves with angular frequencies \( \omega_1 = 52 \) rad/s and \( \omega_2 = 49 \) rad/s. Step 3: Beat Frequency Calculation The beat frequency \( f_{\text{beat}} \) is defined as the absolute difference between the two constituent frequencies:\[ f_{\text{beat}} = |f_1 - f_2| \]First, the angular frequencies must be converted to linear frequencies (in Hz):\[ f_1 = \frac{52}{2\pi} \text{ Hz}, \quad f_2 = \frac{49}{2\pi} \text{ Hz} \]Subsequently, the beat frequency is calculated:\[ f_{\text{beat}} = \frac{|52 - 49|}{2\pi} = \frac{3}{2\pi} \text{ Hz} \] Step 4: Beat Period Calculation The beat period \( T_{\text{beat}} \) is the reciprocal of the beat frequency:\[ T_{\text{beat}} = \frac{1}{f_{\text{beat}}} = \frac{2\pi}{3} \text{ seconds} \]In numerical terms:\[ T_{\text{beat}} \approx \frac{6.2832}{3} \approx 2.0944 \text{ s} \] Step 5: Option Matching The calculated beat period of approximately 2.0944 s is closest to the integer value of 2 s, which corresponds to option (4). Final Answer The correct option is 4.
