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Revision: Wave Motion Physics HSC Science (General) 12th Standard Board Exam Maharashtra State Board

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Theorems and Laws [1]

Prove that the frequency of beats is equal to the difference between the frequencies of the two sound notes giving rise to beats.

Consider two sound waves, having the same amplitude and slightly different frequencies n1 and n2. Let us assume that they arrive in phase at some point x of the medium. The displacement due to each wave at any instant of time at that point is given as

`y_1 = A sin {2pi (n_1t - x/lambda_1)}`

`y_2 = A sin {2pi (n_2t - x/lambda_2)}`

Let us assume for simplicity that the listener is at x = 0.

∴ y1 = A sin (2πn1t)     ...(i)

and y2 = A sin (2πn2t)     ...(ii)

According to the principle of superposition of waves,

y = y1 + y2

∴ y = A sin (2πn1t) + A sin (2πn2t)

By using formula,

sin C + sin D = 2 sin `((C + D)/2) cos ((C − D)/2)`

y = `A[2sin((2pin_1t + 2pi n_2t)/2 )] cos [((2pin_1t - 2pin_2t)/2)]`

y = `2A sin [2pi ((n_1 + n_2)/2)t] cos [2pi ((n_1 - n_2)/2)t]`

∴ y = `R sin [2pi ((n_1 + n_2)/2)t]`

y = R sin (2πnt)     ...(iii)

Where,

R = `2A cos[(2pi(n_1 - n_2))/(2)t]` and n = `(n_1 + n_2)/2`

Equation (iii) is the equation of a progressive wave having frequency `((n_1 + n_2)/2)` and resultant amplitude R.

For waxing,

A = ± 2a

`therefore 2A cos [2pi((n_1 - n_2)/2)t] = +- 2A`

`therefore cos [2pi ((n_1 - n_2)/2)]t = +-( 2A)/(2A)`

`therefore cos [2pi ((n_1 - n_2)/2)]t = +- 1`

This is possible if

`2pi ((n_1 - n_2)/2)t = 0, pi, 2pi, 3pi, ....`

i.e. t = 0, `1/(n_1 - n_2), 2/(n_1 - n_2), 3/(n_1 - n_2), ...`

∴ Period of beat T = `[1/(n_1 - n_2) - 0]`

T = `1/(n_1 - n_2)`

∴ Frequency of beats n = `1/T`

n = n1 − n2

Thus, the frequency of beats is equal to the difference between the frequencies of the two sound notes giving rise to beats.

Important Questions [23]

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