Practice free Waves (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
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 ……….
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 $
Sound waves propagates with a speed of 350 m/s through air and with a speed of 3500 m/s through brass. If a sound wave having frequency 700 hz passes from air to brass, then its wavelength ………….
When, sound waves travel from one medium to another, its frequency does not change. $ \therefore f = { \nu \over \lambda } = constant $ $ \therefore { \nu_a \over \lambda_a} = { \nu_b \over \lambda_b } $ $ \lambda_b = { \nu_b \over \nu_a } \lambda_a = 10 \lambda_a $
A transverse wave is represented by y = ASin (ùt-kx). For what value of its wavelength will the wave velocity be equal to the maximum velocity of the particle taking part in the wave propagation?
Wave velocity = max. velo.of particle $ { \omega \over k } = A \omega $ $ \therefore A = { 1 \over k } = { \lambda \over 2 \pi } $ $ \therefore \lambda = 2 \pi A $
Two monoatomic ideal gases 1 and 2 has molecular weights m1 and m2. Both are kept in two different containers at the same temperature. The ratio of velocity of sound wave in gas 1 and 2 is ……….
Speed of sound in an ideal gas $ \nu = sqrt { \gamma RT \over m } $ $ \therefore { \nu_1 \over \nu_2 } = \sqrt { m_2 \over m_1 } \left ( \therefore v { 1 \over \sqrt m } \right) $
A wire having length L is kept under tension between x = 0 and x = L. In one experiment, the equation of the wave and energy is given by $ y_1 = A sin \left( { \pi x \over L } \right) sin 2 ut $ and $ E_2 $ then
$ E = { 1 \over 2 } m \omega^2 A^2 = {1 \over 2 } m 4 \pi^2 f^2 A^2 $ $ \therefore E \alpha f^2 \Rightarrow { E_1 \over E_2 } = \left( { f_1 \over f_2 } \right)^2 = \left( { f \over 2 f } \right)^2 = { 1 \over 4 } \therefore E_2 = 4 E_1 $
Twenty four tuning forks are arranged in such a way that each fork produces 6 beats/s with the preceding fork. If the frequency of the last tuning fork is double than the first fork, then the frequency of the second tuning fork is ………
Let hte freq. of 1st fork be $f_1$ $ \therefore frequency of 2nd fork = f_1 + 6 = f_1 + 6 (2 -1 ) $ $ \therefore freq. of th 24th fork = f_1 + 6(24 -1 )=f_1 + 138$ Now, freq. of 24th fork = 2 x freq. of 1st fork (given) $ \therefore f_1 + 138 = 2 f_1 \therefore f_1 = 138 Hz $
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