Practice free Waves (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
Equation for a harmonic progressive wave is given by y = Asin ( 15pt + 10px + p/3) where x is in meter and t is in seconds. This wave is ……….
$ on comparing y = A sin ( 15 \pi t + 10 \pi x + { \pi \over 3 }) $ $ with y = A sin ( \omega t + kx + \theta ) $
If the velocity of sound wave in humid air is $v_m$ and that in dry air is $v_d$, then……
At constant pressure density of water vapour is less than dry air. $ \therefore $ with increase in humidityaccording to the equation $ \nu = \sqrt { \gamma p \over \rho } $ the velocity of sound increases.
The ratio of frequencies of two waves travelling through the same mediumis 2:5. The ratio of their wavelengths will be ………
$ f \alpha \lambda^{-1} $ $ \therefore { f_1 \over f_2} = { \lambda_2 \over \lambda_1 } $
If the maximum frequency of a sound wave at room temperature is 20,000 hz then its minimum wavelength will be approximately…….$ ( \nu = 340 ms^{-1} $
From the equation $ v = f \lambda , \lambda_{min} = { \nu \over f_{max} } = 17 mm $ which is nearer to 20 mm
If the equation of a wave in a string having linear mass $ 0.04 kg m^{-1} $ is given by $ y = 0.02 sin \left[ 2 \pi \left( {t \over 0.04 } - { x \over 0.50 } \right) \right] $ , then the tension in the string is……….. N. ( All values are in mks )
On comparing with the wave equation $ y = A sin 2 \pi \left( { t \over T } - { x \over \lambda} \right) we get , T = 0.04 s , \lambda = 0.5 m \Rightarrow \nu = { 25 \over 2 } ms^{-1} $ $ \therefore T = \nu^2 \pi = 6.25 N $
If the equation for a transverse wave is $ y = A sin 2 p \left( {t \over T } - { x \over \lambda} \right) $ , then for what wavelength will the maximum velocity of the particle be double the wave velocity?
Maximum velocity of particle = $A \omega $ $ \therefore wave velocity = f \lambda $ Maximum velocity of particle = $ 2 \times wave velocity$ $ \therefore A \omega = 2 f \lambda \Rightarrow \lambda = \pi A $
Consider two points lying at a distance of 10 mand 15 m from an oscillating source. If the periodic time of oscillation is 0.05 s and the velocity of wave produced is 300 m/s then what will be the phase difference the two points?
Putting values in $ \lambda = \nu T $ $ if phase diff = in the interval \triangle x is \triangle \delta $ then $ \triangle \delta = { 2 \pi \over \lambda } \triangle x = { 2 \pi \over 15} \times (15 -10 ) = { 2 \pi \over 3 } $
A string is divided into three parts having lengths$ l_1$, $l_2$ and $l_3$ each. If the fundamental frequency of these parts are $f_1$, $f_2$ and $f_3$ respectively, then the fundamental frequency of the original string f = ……….
Freq. of a wave in a string $ f \alpha { 1 \over l} $ $ \therefore l = l_1 + l_2 +l_3 $ $ \therefore { 1\over f } = { 1 \over f_1} + { 1 \over f_2 } + { 1 \over f_3} $
Waves produced by two tuning forks are given by $y_1 = 4Sin500pt and y_2 = 2Sin506pt.$ . The number of beats produced per minute is …….
On comparing $y_1 = 4 sin 500 \pi t with y1 = A sin \omega_1 t$ $ we get \omega_1 = 2 \pi f_1 = 500 \pi \Rightarrow f_1 = 250 Hz $ $ Similarly y_2 = 2 sin 506 \pi t $ $ \therefore \omega_2 = 2 \pi f_2 = 506 \pi \Rightarrow f_2 = 253 Hz$ $ \therefore Freq. of beats = f_2 -f_1 = 3 $ $ \therefore No.of beats heard per minute = 3 \times 60 = 180 $
Equation for a progressive harmonic wave is given by y = 8Sin2p( 0.1x – 2t), where x and y are in cm and t is in seconds. What will be the phase difference between two particles of this wave separated by a distance of 2 cm?
$ y = 8 sin 2 \pi ( 0.1 x -2t ) $ $ \therefore y = -8 sin 2 \pi ( 2t - 0.1 x ) comparinf with y = A sin \left( {t \over T } -{ x \over \lambda} \right) $ $ we get { 1 \over \lambda} = 0.1 \Rightarrow \lambda = 10 cm $ $ now path difference between 2 particles \delta = { 2 \pi \over \lambda} .x = kx $ $ \therefore \delta = { 2 \times 180 \times 2 \over 10 } = 72 ^\circ $
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