NEET Physics Questions: Waves

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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 $
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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 $
Two waves are represented by $y_1 = Asinùt$ and $y_2 = aCosùt$. The phase of the first wave, w.r.t. to the second wave is ……….




$ y_1 = a sin \omega t and y_2 = a cos \omega t = a sin ( \omega t + { \pi \over 2 } ) $ $ \therefore 1st wave is lagging behind in phase by { \pi / 2 } $
If the resultant of two waves having amplitude b is b, then the phase difference between the two waves is …….




Here A is the amplitude of resultant wave formed by 2 waves having amplitude $A_1$ and $A_2$ respectively. $ A^2 = A_1^2 + A_2 ^2 + 2 A_1 A_2 cos \theta $ $ Also \theta in the phase A_1 , A_2 $ $ Now putting A_1 = A_2 and A = b , we get $ $ b^2 = 2 b^2 (1 + cos \theta ) $ $ \therefore cos \theta = -{ 1/2} \Rightarrow \theta =120 ^\circ $
If two antinodes and three nodes are formed in a distance of $ 1.21 A ^\circ $ , then the wavelength of the stationary wave is ……….




No explanation available.
If two almost identical waves having frequencies n1 and n2, produced one after the other superposes then the time interval to obtain a beat of maximum intensity is ……..




No. of beats produced per second = $ n_1 - n_2 $ $ \therefore Time interval between 2 consecutive beats = { 1 \over n_1 -n_2 } $
A string of length 70 cmis stretched between two rigid supports. The resonant frequency for this string is found to be 420 hz and 315 hz. If there are no resonant frequencies between these two values, thenwhat would be the minimum resonant frequency of this string?




Let the number of loops obtained for 315Hz and 420Hz n and (n+1) respectively. $ \therefore f_n = nf_1 = 315 $ $ \therefore f_{n+1} = (n+1) f_1 = 420 $ $ \therefore f_{n+1} - f_n = f_1 =105 Hz $