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Two tuning forks have frequencies 450 Hz and 454 Hz respectively. On sounding these forks together, the time interval between successive maximum intensities will be :
The time interval between successive maximum intensities will be
When a tuning fork of frequency 341 is sounded with another tuning fork, six beats per second are heard. When the second tuning fork is loaded with wax and sounded with the first tuning fork, the number of beats is two per second. The natural frequency of the second tuning fork is :
nA = Known frequency = 341 Hz, nB = ?
x = 6 bps, which is decreasing (i.e. x↓) after loading (from 6 to 1 bps)
Unknown tuning fork is loaded so nB↓
Hence nA – nB↓ = x↓ ... (i) → Wrong
nB↓ – nA = x↓ ... (ii) → Correct
⇒ nB = nA + x = 341 + 6 = 347 Hz.
Two tuning forks A and B vibrating simultaneously produce 5 beats. Frequency of B is 512. It is seen that if one arm of A is filed, then the number of beats increases. Frequency of A will be :
After filling frequency increases, so nA decreases (↓). Also it is given that beat frequency increases (i.e., x ↑)
Hence nA↓ – nB = x↑ ... (i) → Correct
nB – nA↑ = x↑ ... (ii) → Wrong
⇒ nA = nB + x = 512 + 5 = 517 Hz.
Beats are produced by two waves given by and . The number of beats heard per second is :
Number of beats per second = n1 ~ n2
⇒ n1 = 1000
and ⇒ n2 = 1004
Number of beats heard per sec = 1004 – 1000 = 4
A tuning fork whose frequency as given by manufacturer is 512 Hz is being tested with an accurate oscillator. It is found that the fork produces a beat of 2 Hz when oscillator reads 514 Hz but produces a beat of 6 Hz when oscillator reads 510 Hz. The actual frequency of the fork is :
The tuning fork whose frequency is being tested produces 2 beats with oscillator at 514 Hz, therefore, frequency of tuning fork may either be 512 or 516. With oscillator frequency 510 it gives 6 beats/sec, therefore frequency of tuning fork may be either 516 or 504.
Therefore, the actual frequency is 516 Hz which gives 2 beats/sec with 514 Hz and 6 beats/sec with 510 Hz.
When a tuning fork A of unknown frequency is sounded with another tuning fork B of frequency 256 Hz, then 3 beats per second are observed. After that A is loaded with wax and sounded, the again 3 beats per second are observed. The frequency of the tuning fork A is :
It is given that :
nA = Unknown frequency = ?
nB = Known frequency = 256 Hz
x = 3 bps, which remains same after loading
Unknown tuning fork A is loaded so nA↓
Hence nA↓ – nB = x ... (i) → Correct
nB – nA↓ = x ... (ii) → Wrong
⇒ nA = nB + x = 256 + 3 = 259 Hz.
When two sound waves are superimposed, beats are produced when they have :
For producing beats, their must be small difference in frequency.
A tuning fork A of frequency 200 Hz is sounded with fork B, the number of beats per second is 5. By putting some wax on A, the number of beats increases to 8. The frequency of fork B is :
nA↓ – nB = x↑ ... (i) → Wrong
nB – nA↓ = x↑ ... (ii) → Correct
⇒ nB = nA + x = 200 + 5 = 205 Hz.
Two tuning forks have frequencies 380 and 384 Hz respectively. When they are sounded together, they produce 4 beats. After hearing the maximum sound, how long will it take to hear the minimum sound?
Beat period .
Hence minimum time interval between maxima and minima
A couple of tuning forks produces 2 beats in the time interval of 0.4 seconds. So the beat frequency is :
Beat frequency =
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