A source producing the sound of frequency 170 Hz is approaching a stationary observer with a velocity of 17 ms–1. The apparent change in the wavelength of sound heard by the observer is (speed of sound in air = 340 ms–1)
Now
⇒
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A source producing the sound of frequency 170 Hz is approaching a stationary observer with a velocity of 17 ms–1. The apparent change in the wavelength of sound heard by the observer is (speed of sound in air = 340 ms–1)
Now
⇒
An observer moves towards a stationary source of sound with a speed 1/5th of the speed of sound. The wavelength and frequency of the sound emitted are λ and f respectively. The apparent frequency and wavelength recorded by the observer are respectively :
and since the source is stationary, so wavelength remains unchanged for observer.
The equation of displacement of two waves are given as ; . Then what is the ratio of their amplitudes ?
...(i)
and
... (ii)
(∵ sin(A + B) = sinA cosB + cosA sinB)
Comparing equation (i) and (ii) we get ratio of amplitude 1 : 1.
Consider ten identical sources of sound all giving the same frequency but having phase angles which are random. If the average intensity of each source is I0, the average of resultant intensity I due to all these ten sources will be :
In case of interference of two waves resultant intensity
If Ï• varies randomly with time, so
⇒
For n identical waves,
Here
41 forks are so arranged that each produces 5 beats per sec when sounded with its near fork. If the frequency of the last fork is double the frequency of the first fork, then the frequencies of the first and last fork are respectively :
Similar to previous question
nFirst = nFirst + (N – 1)x
2n = n + (41 – 1) × 5
⇒ nFirst = 200 Hz and nLast = 400 Hz
Two identical wires have the same fundamental frequency of 400 Hz when kept under the same tension. If the tension in one wire is increased by 2%, the number of beats produced will be :
⇒
Beat frequency
16 tunning forks are arranged in the order of increasing frequencies. Any two successive forks give 8 beats per sec when sounded together. If the frequency of the last fork is twice the first, then the frequency of the first fork is
Using nLast = nFirst + (N – 1)x
⇒ 2n = n + (16 – 1) × 8 ⇒ n = 120 Hz
The frequency of a stretched uniform wire under tension is in resonance with the fundamental frequency of a closed tube. If the tension in the wire is increased by 8 N, it is in resonance with the first overtone of the closed tube. The initial tension in the wire is
According to problem
…..(i)
and ..…(ii)
Dividing equation (i) and (ii),
A metal wire of linear mass density of 9.8 g/m is stretched with a tension of 10 kg weight between two rigid supports 1 metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency n. The frequency n of the alternating source is :
In condition of resonance, frequency of a.c. will be equal to natural frequency of wire
An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency f1. Now it is closed at one end. If the frequency of the tuning fork is increased slowly from f1 , then again a resonance is obtained with a frequency f2. If in this case the pipe vibrates in harmonic, then -
Open pipe resonance frequency
Closed pipe resonance frequency
(where n is odd and )
∴ n = 5
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