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Q.Prove that the number of beats heard per second is equal to the difference in frequencies of two sound sources. OR Obtain an expression for the time-period of a simple pendulum.

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2022Subjective· 5mImportance★★★★★
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Superposing two waves of close frequencies f₁ and f₂ produces beats at a rate equal to |f₁−f₂| per second.

Setup. Consider two sound waves of equal amplitude aa but slightly different frequencies f1f_1 and f2f_2 (f1>f2f_1>f_2), travelling in the same direction, observed at a fixed point:

y1=asin⁡(2πf1t),y2=asin⁡(2πf2t)y_1 = a\sin(2\pi f_1t), \qquad y_2 = a\sin(2\pi f_2t)

Superposition. By the principle of superposition, the resultant displacement is:

y=y1+y2=a[sin⁡(2πf1t)+sin⁡(2πf2t)]y = y_1+y_2 = a[\sin(2\pi f_1t)+\sin(2\pi f_2t)]

Using the sum-to-product identity sin⁡C+sin⁡D=2sin⁡ ⁣(C+D2)cos⁡ ⁣(C−D2)\sin C+\sin D = 2\sin\!\left(\frac{C+D}{2}\right)\cos\!\left(\frac{C-D}{2}\right):

y=2acos⁡ ⁣(2πf1−f22t)sin⁡ ⁣(2πf1+f22t)y = 2a\cos\!\left(2\pi\frac{f_1-f_2}{2}t\right)\sin\!\left(2\pi\frac{f_1+f_2}{2}t\right)

This represents a wave of frequency fˉ=f1+f22\bar f = \dfrac{f_1+f_2}{2} (the average, audible tone), whose amplitude itself varies slowly with time as:

A(t)=2acos⁡ ⁣(2πf1−f22t)A(t) = 2a\cos\!\left(2\pi\frac{f_1-f_2}{2}t\right)

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