NEET Practice Questions (MCQs) with Answers & Solutions

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 $

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 } $

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} $

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 $

$ 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 $

$ 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 } $

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 $

$ y = sin^2 \omega t = {1 - cos 2 \omega t \over 2 } = {1 \over 2 } - {1 \over 2 } cos 2 \omega t .....(1) $ $ \therefore \nu = { 1 \over 2 } 2 \omega sin ( 2 \omega t ) = \omega sin 2 \omega t $ $ \therefore a = 2 \omega^2 cos 2 \omega t $ $ = 2 \times 2 \omega^2 \left( {1 \over 2} - Y \right) \{ From eqn (1) \} $ $ = - 4 \omega^2 \left( { 1 \over 2} - y \right) $ $ \therefore a \alpha - y \{ \therefore SHM \} $ $ Now , { 2 \pi \over T} = 2 \omega \Rightarrow T = { \pi \over \omega } $

No. of beats produced per second = $ n_1 - n_2 $ $ \therefore Time interval between 2 consecutive beats = { 1 \over n_1 -n_2 } $