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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- A transverse wave is produced on a stretched string 0.9 m long and fixed at its ends. Find the speed of the transverse wave, when the string vibrates while emitting second overtone of frequency
- Explain the Reflection of Transverse and Longitudinal Waves from a Denser Medium and a Rared Medium.
- Derive an Expression for Frequency in Fundamental Mode
- A Set of 48 Tuning Forks is Arranged in a Series of Descending Frequencies Such that Each Fork Gives 4 Beats per Second with Preceding One.
- A Sonometer Wire is in Unison with a Tuning Fork of Frequency 125 Hz When It is Stretched by a Weight. When the Weight is Completely Immersed in Water, 8 Beats Are Heard per Second.
- A train blows a whistle of frequency 640 Hz in air. Find the difference in apparent frequencies of the whistle for a stationary observer, when the train moves towards and away from the
- Let N1 and N2 Be the Two Slightly Different Frequencies of Two Sound Waves. the Time Interval Between Waxing and Immediate Next Waning is
- A Set of 12 Tuning Forks is Arranged in Order of Increasing Frequencies. Each Fork Produces ‘Y’ Beats per Second with the Previous One. the Last is an Octave of the First.
- The Working of Radar is Based on
- Doppler Effect is Not Applicable When
- Apparent Frequency of the Sound Heard by a Listener is Less than the Actual Frequency of Sound Emitted by Source. in this Case
- State Any Four Applications Of Doppler Effect
- In Doppler Effect of Light, the Term “Red Shift” is Used for
- Find the Velocity of Sound in the Air and Frequency of the Tuning Fork
- Prove that the frequency of beats is equal to the difference between the frequencies of the two sound notes giving rise to beats.
- Two tuning forks of frequencies 320 Hz and 340 Hz are sounded together to produce a sound wave. The velocity of sound in air is 326.4 m/s. Calculate the difference in wavelengths of these waves.
