The displacement of the interfering light waves are and . What is the amplitude of the resultant wave :
Since,
⇒
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The displacement of the interfering light waves are and . What is the amplitude of the resultant wave :
Since,
⇒
Two waves are represented by and . What will be their resultant amplitude :
Putting a1 = a2 = a and , we get
The amplitude of a wave represented by displacement equation will be
Here phase difference =
∴ The resultant amplitude
=
A tuning fork sounded together with a tuning fork of frequency 256 Hz emits two beats. On loading the tuning fork of frequency 256 Hz with wax, the number of beats heard are 1 per second. The frequency of the other tuning fork is :
nA = Known frequency = 256, nB = ?
x = 2 bps, which is decreasing after loading (i.e. x↓) known tuning fork is loaded so nA↓
Hence nA↓ – nB = x↓ ... (i) → Correct
nB – nA↓ = x↓ ... (ii) → Wrong
⇒ nB = nA – x = 256 – 2 = 254 Hz.
If two tuning forks A and B are sounded together, they produce 4 beats per second. A is then slightly loaded with wax, they produce 2 beats when sounded again. The frequency of A is 256. The frequency of B will be :
nA = Known frequency = 256 Hz, nB = ?
x = 4 bps, which is decreasing after loading (i.e. x↓) also known tuning fork is loaded so nA↓
Hence nA↓ – nB = x↓ ... (i) → Correct
nB – nA↓ = x↓ ... (ii) → Wrong
⇒ nB = nA – x = 256 – 4 = 252 Hz.
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.
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