Revision: Oscillations and Waves JEE Main Oscillations and Waves

The number of oscillations produced by the particle per second is called frequency. It is denoted by f. SI unit for frequency is s⁻¹ or hertz (Hz).

Mathematically, frequency is related to the time period by f = `1/"T"`

Time period: The time period is defined as the time taken by a particle to complete one oscillation. It is usually denoted by T. For one complete revolution, the time taken is t = T, therefore,

`ω"T"` = 2π ⇒ T = `(2π)/ω`

The apparent change in the frequency of sound heard by a listener, due to relative motion between the source of sound and the listener is called Doppler effect in sound.

When the source and the observer are in relative motion with respect to each other and to the medium in which sound propagates, the frequency of the sound wave observed is different from the frequency of the source. This phenomenon is called Doppler Effect.

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.