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 nearest integer)

Solution & Explanation

### Related Formula The product of cosines can be transformed into a sum using the trigonometric identity: cos A cos B = frac12 [cos(A + B) + cos(A - B)] The beat frequency f_textbeat is given by: f_textbeat = |f_1 - f_2| = left| fracomega_1 - omega_22pi right| The beat period T_textbeat is: T_textbeat = frac1f_textbeat ### Core Logic Rewrite the superposition equation: x = a cos(1.5t) cos(50.5t) Apply the identity with A = 50.5t and B = 1.5t: x = fraca2 [cos(52t) + cos(49t)] Here, the two component frequencies are: omega_1 = 52 mathrm~rad/s implies f_1 = frac522pi omega_2 = 49 mathrm~rad/s implies f_2 = frac492pi Calculate the beat frequency: f_textbeat = f_1 - f_2 = frac52 - 492pi = frac32pi mathrm~Hz ### Step 1: Calculate Beat Period The time period of beats is: T_textbeat = frac1f_textbeat = frac2pi3 approx frac2 times 3.143 = 2.09 mathrm~s Rounding to the nearest integer gives 2 mathrm~s. ### Pattern Recognition Sees: product of two cosines with significantly different coefficients omega_1 and omega_2. Shortcut: The beat period is simply 2pi divided by the difference between the two component frequencies, where the component frequencies are (omega_textaverage pm omega_textenvelope). The difference is 2 times omega_textenvelope = 2 times 1.5 = 3 mathrm~rad/s. Thus, T = 2pi / 3 approx 2 mathrm~s. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

Reference Study Guides

More Waves Previous-Year Questions — Page 4

Q45 jee_main_2024_31_jan_morning Organ Pipes
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60mathrm~cm, the length of the closed pipe will be:
  • A. 60mathrm~cm
  • B. 45mathrm~cm
  • C. 30mathrm~cm
  • D. 15mathrm~cm

Solution

### Related Formula f_textclosed, fundamental = fracv4L_c f_textopen, 1st overtone = frac2v2L_o ### Core Logic
Organ Pipes diagram for Q45 - JEE Main 2024 Morning
Organ Pipes diagram for Q45 - JEE Main 2024 Morning
Organ Pipes diagram for Q45 - JEE Main 2024 Morning
Organ Pipes diagram for Q45 - JEE Main 2024 Morning
For a closed organ pipe, the fundamental frequency (1st harmonic) is: f_1 = fracvlambda = fracv4L_1 where L_1 is the length of the closed pipe. For an open organ pipe, the first overtone (2nd harmonic) is: f_2 = frac2v2L_2 = fracvL_2 where L_2 is the length of the open pipe (L_2 = 60mathrm\,cm). ### Step 2: Equating Frequencies Given f_1 = f_2: fracv4L_1 = fracvL_2 L_2 = 4L_1 60 = 4 times L_1 L_1 = 15mathrm\,cm ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

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