Explanation
Increase in tension of string increases its frequency. If the original frequency of B(fB) were greater than that of A (fA), further the increase in fB should have resulted in increase in the beat frequency. But the beat frequency is found to decrease. This shows that fA- fB = 5 Hz and
fA=427 Hz, we get
fB = 422 Hz
The was decreases the frequency of unknown fork. The possible unknown frequencies are, (288+4) Hz and (288-4) Hz. Wax reduces 284 Hz and so beats should increases. It is not given in the question. This frequency is ruled out. Wax reduces 292 Hz and so beats should decrease. It is given that the beats decrease from 2 to 4. Hence the unknown fork has frequency 292 Hz.
When temperature increases, l increases. Hence frequency decreases.
In a longitudinal wave, pressure is maximum where displacement is minimum. Therefore pressure and displacement variations are 180o out of phase
Frequency of tuning fork f1 = 480 Hz. Number of beats s – 1, n = 10
Frequency of string f2 = (480 + 10)Hz. A slight increase in tension increase f2
f2 = 480 – 10 = 470 Hz.
(c) is the correct choice because its value is finite at all times.
As sin (90 ± θ) = cos θ ∴ The phase difference between the two waves is Π/2
When other end of pipe is opened, its fundamental frequency becomes 200Hz. The overtone have frequencies 400, 600, 800 Hz..
As the source is moving perpendicular to straight line joining the observer and source, (as if moving along a circle), apparent frequency is not affected n1 = 0
At a displacement antinode, a pressure node is present. Since pressure does not change at its node, nor does density.
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