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A wave travelling along a string is described by y = 0.005Sin(40x – 2t) in SI units. The wavelength and frequency of the wave are………
$ From the phase angle ( 40 -2t ) , we get k =40 \, OR\, { 2 \pi \over \lambda } = 40 \Rightarrow \lambda = { \pi \over 20 } $ $ and \omega = 2 \, OR \, 2 \pi f = 2 \Rightarrow f = \pi^{-1} Hz $
Two sitar strings A and B playing the note “Dha†are slightly out of time and produce beats of frequency 5 hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 hz. What is the original frequency of B if the frequency of A is 427 hz?
Increase in tension of string increases its frequency. If the original frequency of $B(f_B)$ were greater than that of $A(f_A)$, further the increase in $f_B$ should have resulted in increase in the beat frequency. But the beat frequency is found to decrease. This shows that $f_A-f_B = 5 Hz$ and $f_A=427 Hz$, we get $f_B = 422 Hz$
A rocket is moving at a speed of 130 m/s towards a stationary target. While moving, it emits a wave of frequency 800 hz. Calculate the frequency of the sound as detected by the target. ( Speed of wave = 330 m/s)
$ f_L = \left( { \nu - \nu_L \over \nu - \nu_s } \right) f_s = \left[ { 330 - 0 \over 330 -130 } \right] \times 800 = 1320 Hz $
Length of a steel wire is 11 m and its mass is 2.2 kg. What should be the tension in the wire so that the speed of a transverse wave in it is equal to the speed of sound in dry air at $20 ^\circ C $ temperature?
$ \nu = \sqrt T \Rightarrow T = \mu \nu^2 = { M \over L } \nu^2 = { 2.2 \over 11 } \times (340)^2 $ $ \therefore T = 2.31 \times 10^4 N $
A wire stretched between two rigid supports vibrates with a frequency of 45 hz. If the mass of the wire is $3.5 \times 10^{ – 2} kg $ and its linear mass density is $4.0 \times 10^{- 2} kg/m$, what will be the tension in the wire ?
$ f = { 1 \over 21 } \sqrt T \Rightarrow T = f^2 4 l^2 $ $ \therefore T = 4 f^2 ( M)^2 = 4 f^2 M^2 = 248 N$
Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube A and B is ……..
$ In tube A , \lambda_A = 2 l $ $ In tube B , \lambda _B = 4 l $ $ \therefore \nu_A = { \nu \over \lambda_A } = { \nu \over 21} $ $ \therefore \nu_B = { \nu \over \lambda_B } = { \nu \over 41 } \Rightarrow { \nu_A \over \nu_B } ={ 2 \over 1 } $
A tuning fork arrangement produces 4 beats/second with one fork of frequency 288 hz. A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is……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. consider option (a)
A wave y = aSin(ùt – kx ) on a string meets with another wave producing a node at x= 0. Then the equation of the unknown wave is ………
Stationary wave : Y = a sin (wt-kx) + a sin(wt+kx) When x = 0, $Y \neq 0$. The option is not acceptable consider option (b) stationary wave : Y = a sin (wt-kx) - a sin(wt+kx) At x = 0, Y = 0. This option holds good. Option (c) gives Y = 2a sin(wt - kx) At x = 0, $Y \neq 0$ Option (d) gives Y = 0. Hence option (b) holds good
A tuning fork of known frequency 256 hz makes 5 beats per second with the vibrating string of a piano. The beats frequency decreases to 2 beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increase in the tension was hz.
The possible frequency of piano are (256 + 5)Hz and (256 - 5)Hz. For a piano string $ \nu = { 1 \over 2l } \sqrt {T} $ When tension T increases v increases. (i) If 261 Hz increases, beats / second increase. This is not given. (ii) If 251 Hz increases due to tension, beats / second decrease. This is given.
An observer moves towards a stationary source of sound with a velocity one – fifth the velocity of sound. What is the percentage increase in the apparent frequency?
$ By doppler's effect , { f_L \over f_S} = \left( { \nu + \nu_L \over \nu + \nu_S} \right) $ $ \therefore { f_L \over f_S} = { \nu + \nu_L \over \nu } = { \nu + \nu/5 \over \nu } = { 6 \over 5 } $ $ \therefore Fractional increase = { f_L - f_S \over f_S } = { f_L \over f_S } -1 = { 6 \over 5} -1 = {1 \over 5} $ $ \therefore Percentage increase = { 100 \over 5} = 20 \% $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.